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0$).","code":"B"}],"finalAnswers":["$$f(x) \\geq 0$","$$y \\geq 0$","$$[0, \\infty)$"],"generatedAt":"2026-05-06T14:04:54.008Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"algebraic","label":"Algebraic Method","steps":[{"type":"text","order":0,"title":"Understand the goal","content":"To find the range of the function $f(x)$, we need to determine the set of all possible output values (y-values) it can produce, given the domain constraint $x \\geq 2$. We will achieve this by rewriting the quadratic expression as a perfect square.","highlights":[{"text":"State the range of f.","location":"specification"},{"text":"f(x) = x^2 - 4x + 4, \\quad x \\in \\mathbb{R}, x \\geq 2","location":"specification"}]},{"type":"text","order":1,"title":"Factor into perfect square","content":"First, we factor the quadratic expression $x^2 - 4x + 4$ into a perfect square:\n\n$$f(x) = (x - 2)^2$$\n\nThis step earns the first method mark (**M1**)."},{"type":"text","order":2,"title":"Apply the domain constraint","content":"Now we use the given domain constraint, $x \\geq 2$, to find the possible values of $f(x)$:\n\n1. Start with the domain constraint:\n $$x \\geq 2$$\n2. Subtract $2$ from both sides:\n $$x - 2 \\geq 0$$\n3. Square both sides (since the terms are non-negative, the inequality direction is preserved):\n $$(x - 2)^2 \\geq 0$$"},{"type":"text","order":3,"title":"State the final range","content":"Since $f(x) = (x - 2)^2$ and $(x - 2)^2 \\geq 0$ for all $x \\geq 2$, the minimum value of $f(x)$ is $0$ (which occurs when $x = 2$). As $x$ increases, $(x-2)^2$ increases without limit.\n\nTherefore, the range of $f$ is:\n\n$$f(x) \\geq 0$$\n\nThis completes the second accuracy mark (**A1**)."}]},{"id":"graphical","label":"Graphical Approach","steps":[{"type":"text","order":0,"title":"Analyze the function","content":"To find the range of $f(x) = x^2 - 4x + 4$ for $x \req 2$, we can analyze its graph. First, let's write the quadratic expression in vertex form by completing the square or factoring it.\n\nObserve that the quadratic is a perfect square trinomial:\n$$f(x) = (x - 2)^2$$\n\nThis form allows us to easily identify the key features of the parabola.","highlights":[{"text":"f(x) = x^2 - 4x + 4, \\quad x \\in \\mathbb{R}, x \\geq 2","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(x) = (x - \\htmlData{anchor=vertex-x}{\\color{@sky}{2}})^2 + \\htmlData{anchor=vertex-y}{\\color{@amber}{0}}$"},{"latex":"$$\\text{Vertex: } (\\htmlData{anchor=v-x}{\\color{@sky}{2}}, \\htmlData{anchor=v-y}{\\color{@amber}{0}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"vertex-x"},{"color":"amber","style":"text-color","anchor":"vertex-y"}],"connections":[{"to":"v-x","from":"vertex-x","color":"sky","label":"x-coord","style":"solid"},{"to":"v-y","from":"vertex-y","color":"amber","label":"y-coord","style":"solid"}]},"order":1,"title":"Identify the vertex","description":"From the vertex form, the vertex of the parabola is located at $(2, 0)$."},{"type":"text","order":2,"title":"Determine the shape and domain","content":"Let's consider the shape and the restricted domain of the function:\n- Since the coefficient of the $x^2$ term is positive ($+1$), the parabola opens upwards.\n- The domain of the function is restricted to $x \ngeq 2$.\n\nBecause the domain starts exactly at the $x$-coordinate of the vertex ($x = 2$) and goes to positive infinity ($x \\to \\infty$), we are only looking at the right-hand branch of the parabola.","highlights":[{"text":"x \\geq 2","location":"specification"}]},{"type":"text","order":3,"title":"Find the range","content":"Since the graph starts at the vertex $(2, 0)$ and curves upwards as $x$ increases, the minimum value of $f(x)$ is the $y$-value of the vertex, which is $0$.\n\nAs $x$ increases towards infinity, $f(x)$ also increases towards infinity.\n\nTherefore, the range of the function is all values greater than or equal to $0$:\n$$f(x) \\geq 0 \\quad \\text{or} \\quad [0, \\infty)$$\n\nWriting this down as $f(x) \\geq 0$ (or $y \\geq 0$) will earn you full marks."}]}],"generatedAt":"2026-06-11T03:08:13.658Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"b1cba5e7-cfe1-4c8b-9931-b27baf952724","label":"b","content":"Sketch the graphs of $f(x)$ and $f^{-1}(x)$ on the same diagram.","markscheme":"- Sketch of $f(x)$ starting at $(2, 0)$ and passing through $(3, 1)$ B1\n- Sketch of $f^{-1}(x)$ starting at $(0, 2)$ and passing through $(1, 3)$ B1\n- Reflection in the line $y=x$ shown B1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Correct shape for $f(x)$ starting at $(2, 0)$.","code":"B"},{"id":"p2","marks":1,"description":"Correct shape for $f^{-1}(x)$ starting at $(0, 2)$.","code":"B"},{"id":"p3","marks":1,"description":"Indication of reflection in the line $y=x$.","code":"B"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:04:54.008Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"geometric-reflection","label":"Geometric Reflection","steps":[{"type":"text","order":0,"title":"Analyze the function","content":"First, let's write the given function in its completed square form to easily identify its shape and key points:\n\n$$f(x) = x^2 - 4x + 4 = (x-2)^2$$"},{"type":"text","order":1,"title":"Plot key points of f(x)","content":"Since the domain of $f(x)$ is $x \\ge 2$, we can determine a few key points on its curve:\n\n* **Starting point (vertex):** When $x = 2$, $f(2) = (2-2)^2 = 0$, giving the point $$(2, 0)$$\n* **Second point:** When $x = 3$, $f(3) = (3-2)^2 = 1$, giving the point $$(3, 1)$$\n* **Third point:** When $x = 4$, $f(4) = (4-2)^2 = 4$, giving the point $$(4, 4)$$\n\nDraw a smooth curve starting at $(2,0)$ and passing upwards through $(3,1)$ and $(4,4)$ to sketch $f(x)$."},{"type":"hyper_latex","block":{"nodes":[{"text":"An inverse function is a geometric reflection of the original function across the line $y=x$. This means we swap the $x$ and $y$ coordinates of each key point:"},{"latex":"$$\\htmlData{anchor=pt1}{(2, 0)} \\quad \\xrightarrow{\\text{reflect across } y=x} \\quad \\htmlData{anchor=pt1-ref}{(0, 2)}$"},{"latex":"$$\\htmlData{anchor=pt2}{(3, 1)} \\quad \\xrightarrow{\\text{reflect across } y=x} \\quad \\htmlData{anchor=pt2-ref}{(1, 3)}$"},{"latex":"$$\\htmlData{anchor=pt3}{(4, 4)} \\quad \\xrightarrow{\\text{reflect across } y=x} \\quad \\htmlData{anchor=pt3-ref}{(4, 4)}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"pt1"},{"color":"sky","style":"text-color","anchor":"pt1-ref"},{"color":"pink","style":"text-color","anchor":"pt2"},{"color":"pink","style":"text-color","anchor":"pt2-ref"}],"connections":[{"to":"pt1-ref","from":"pt1","color":"sky","label":"swap x and y","style":"solid"},{"to":"pt2-ref","from":"pt2","color":"pink","label":"swap x and y","style":"solid"}]},"order":2,"title":"Reflect points across y=x","description":"Each point $(a,b)$ on the graph of $f(x)$ corresponds to a point $(b,a)$ on the graph of $f^{-1}(x)$."},{"type":"text","order":3,"title":"Sketch both graphs","content":"To complete the sketch on your axes:\n\n1. Draw the line of symmetry $$y=x$$ as a dashed diagonal line.\n2. Plot the reflected points $(0, 2)$, $(1, 3)$, and $(4, 4)$.\n3. Draw a smooth curve starting from $(0, 2)$ on the $y$-axis and curving through $(1, 3)$ and $(4, 4)$ to represent $f^{-1}(x)$.\n\n### Markscheme Breakdown\n* **B1:** Curve for $f(x)$ starts at $(2,0)$ and passes through $(3,1)$.\n* **B1:** Curve for $f^{-1}(x)$ starts at $(0,2)$ and passes through $(1,3)$.\n* **B1:** The line of reflection $y=x$ is clearly shown, with both curves symmetric about it."}]},{"id":"algebraic-inverse","label":"Algebraic Derivation and Plotting","steps":[{"type":"text","order":0,"title":"Analyze the function f(x)","content":"To sketch the graphs, we first examine the original function and its key features:\n\n$$f(x) = x^2 - 4x + 4, \\quad x \\geq 2$$\n\nThis can be rewritten by completing the square or factoring as a perfect square:\n\n$$f(x) = (x - 2)^2$$\n\nSince the domain is restricted to $x \\geq 2$, let's find two key coordinate points to plot this curve:\n- **Starting point**: When $x = 2$, $f(2) = (2 - 2)^2 = 0$. This gives the point $(2, 0)$.\n- **Another point**: When $x = 3$, $f(3) = (3 - 2)^2 = 1$. This gives the point $(3, 1)$.\n\nThe graph of $f(x)$ starts at $(2, 0)$ and curves upwards as a parabola through $(3, 1)$.","highlights":[{"text":"f(x) = x^2 - 4x + 4, \\quad x \\in \\mathbb{R}, x \\geq 2","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=f-eq}{(x - 2)^2}$"},{"latex":"$$\\sqrt{y} = \\htmlData{anchor=f-sqrt}{x - 2} \\quad \\text{since } x \\geq 2$"},{"latex":"$$x = \\htmlData{anchor=inv-y}{\\sqrt{y} + 2}$"},{"latex":"$$f^{-1}(x) = \\htmlData{anchor=inv-final}{\\sqrt{x} + 2}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"f-eq"},{"color":"amber","style":"box","anchor":"inv-final"}],"connections":[{"to":"f-sqrt","from":"f-eq","color":"sky","label":"square root","style":"solid"},{"to":"inv-final","from":"inv-y","color":"amber","label":"swap x and y","style":"solid"}]},"order":1,"title":"Find the inverse function","description":"We set $y = f(x)$ and solve for $x$ to find the algebraic equation of the inverse function $f^{-1}(x)$."},{"type":"text","order":2,"title":"Analyze the inverse function","content":"For the inverse function $f^{-1}(x) = \\sqrt{x} + 2$, the domain is the range of $f(x)$. Since $(x-2)^2 \\geq 0$, the domain of $f^{-1}(x)$ is $x \\geq 0$.\n\nLet's find key coordinate points for $f^{-1}(x)$:\n- **Starting point**: When $x = 0$, $f^{-1}(0) = \\sqrt{0} + 2 = 2$. This gives the point $(0, 2)$.\n- **Another point**: When $x = 1$, $f^{-1}(1) = \\sqrt{1} + 2 = 3$. This gives the point $(1, 3)$.\n\nNotice that these points are the exact reflections of the points of $f(x)$ across the line of symmetry $y = x$:\n- $(2, 0) \\leftrightarrow (0, 2)$\n- $(3, 1) \\leftrightarrow (1, 3)$"},{"type":"text","order":3,"title":"Sketch the curves","content":"Now, sketch both curves on the same Cartesian coordinate system:\n\n1. Draw the line of reflection $y = x$ as a dashed line.\n2. Plot and label $f(x)$ starting at $(2, 0)$, curving up through $(3, 1)$.\n3. Plot and label $f^{-1}(x)$ starting at $(0, 2)$, curving to the right through $(1, 3)$.\n4. Ensure the curves are symmetrical to each other across the line $y=x$.\n\n**Markscheme Tips:**\n- You get **1 mark** for correctly sketching $f(x)$ from $(2, 0)$ passing through $(3, 1)$.\n- You get **1 mark** for correctly sketching $f^{-1}(x)$ from $(0, 2)$ passing through $(1, 3)$.\n- You get **1 mark** for showing the line of symmetry $y=x$.","calculator":[{"gdc":{"instructions":"1. Go to Graph mode.\n2. Define \\\\f1(x) = (x-2)^2\\\\ with domain constraint \\\\x \\ge 2\\\\.\n3. Define \\\\f2(x) = \\sqrt{x} + 2\\\\.\n4. Define \\\\f3(x) = x\\\\.\n5. Set the window aspect ratio to square to visualize symmetry."},"ti84":{"instructions":"1. Press [Y=].\n2. Enter \\\\Y_1 = (X - 2)^2 / (X \\ge 2)\\\\ (use the logic test menu [2nd][MATH] for the inequality).\n3. Enter \\\\Y_2 = \\sqrt{X} + 2\\\\.\n4. Enter \\\\Y_3 = X\\\\.\n5. Press [ZOOM] [5] (ZSquare) to ensure equal scaling on both axes to see the reflection symmetry properly."},"desmos":{"instructions":"1. In the first input bar, type: \\\\f(x) = (x-2)^2 \\{x \\ge 2\\\\}.\n2. In the second input bar, type: \\\\g(x) = \\sqrt{x} + 2\\\\.\n3. In the third input bar, type: \\\\y = x\\\\ (change style to dashed in the settings)."},"description":"Using a graphing utility helps you visualize the curves and verify their symmetry across $y = x$."}]}]}],"generatedAt":"2026-06-11T03:08:34.960Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"2161411d-0a4c-470d-901d-2f5e661b087c","label":"c","content":"Find an expression for $f^{-1}(x)$ and state its domain.","markscheme":"- $y = (x-2)^2 \\Rightarrow \\sqrt{y} = x-2$ M1\n- $x = \\sqrt{y} + 2$ M1\n- $f^{-1}(x) = \\sqrt{x} + 2$ A1\n- Domain: $x \\geq 0$ B1","marks":4,"order":2,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Correctly identifies $f(x) = (x-2)^2$ and takes the square root.","code":"M"},{"id":"p2","marks":1,"description":"Rearranges to make $x$ the subject.","code":"M"},{"id":"p3","marks":1,"description":"Correct expression for $f^{-1}(x)$.","code":"A"},{"id":"p4","marks":1,"description":"States domain as $x \\geq 0$.","code":"B"}],"finalAnswers":["$$f^{-1}(x) = \\sqrt{x} + 2, x \\geq 0$"],"generatedAt":"2026-05-06T14:04:54.008Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"completing_square","label":"Completing the Square","steps":[{"type":"text","order":0,"title":"Complete the square","content":"To find the inverse function, we first express $f(x)$ in a completed square form.\n\nLooking at the quadratic expression:\n$$f(x) = x^2 - 4x + 4$$\n\nThis is a perfect square trinomial, which factorises directly to:\n$$y = (x - 2)^2$$\n\nWriting this as $y = (x - 2)^2$ earns you the first mark **(M1)**.","highlights":[{"text":"f(x) = x^2 - 4x + 4, \\quad x \\in \\mathbb{R}, x \\geq 2","location":"specification"}]},{"type":"text","order":1,"title":"Swap variables","content":"To find the inverse function, we swap the variables $x$ and $y$:\n$$x = (y - 2)^2$$"},{"type":"text","order":2,"title":"Solve for y","content":"Now we rearrange to solve for $y$ by taking the square root of both sides:\n$$\\sqrt{x} = \\pm(y - 2)$$\n\nSince the domain of our original function is restricted to $x \\geq 2$, the range of our inverse function $f^{-1}(x)$ (which corresponds to $y$) must also be $y \\geq 2$. Therefore, we must choose the positive square root:\n$$\\sqrt{x} = y - 2$$\n\nRearranging to make $y$ the subject:\n$$y = \\sqrt{x} + 2$$\n\nThis rearrangement and selection of the positive root earns you the second method mark **(M1)**.","highlights":[{"text":"x \\geq 2","location":"specification"}]},{"type":"text","order":3,"title":"State the inverse function","content":"We replace $y$ with $f^{-1}(x)$ to get our final inverse expression:\n$$f^{-1}(x) = \\sqrt{x} + 2$$\n\nThis correct expression earns the accuracy mark **(A1)**."},{"type":"text","order":4,"title":"Determine the domain","content":"The domain of the inverse function $f^{-1}(x)$ is equal to the range of the original function $f(x)$.\n\nGiven the domain of $f(x)$ is $x \\geq 2$, we can find the range of $f(x)$ by substituting the boundary value:\n- When $x = 2$, $f(2) = (2 - 2)^2 = 0$\n- For all $x \\geq 2$, $f(x) \\geq 0$\n\nTherefore, the range of $f(x)$ is $f(x) \\geq 0$.\n\nThis means the domain of $f^{-1}(x)$ is:\n$$x \\geq 0$$\n\nStating this domain correctly earns the final mark **(B1)**."}]},{"id":"quadratic_formula","label":"Quadratic Formula","steps":[{"type":"text","order":0,"title":"Set equal to y","content":"To find the inverse function, we first set the function expression equal to $y$:\n\n$$y = x^2 - 4x + 4$$"},{"type":"text","order":1,"title":"Rearrange to quadratic form","content":"Next, we rearrange the equation into a standard quadratic form $ax^2 + bx + c = 0$ with respect to $x$ by subtracting $y$ from both sides:\n\n$$x^2 - 4x + (4 - y) = 0$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2 - 4x + (\\htmlData{anchor=c-val}{\\color{@sky}{4 - y}}) = 0$"},{"latex":"$$x = \\frac{-(-4) \\pm \\sqrt{(-4)^2 - 4(1)(\\htmlData{anchor=c-sub}{\\color{@sky}{4 - y}})}}{2(1)}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"c-val"},{"color":"sky","style":"text-color","anchor":"c-sub"}],"connections":[{"to":"c-sub","from":"c-val","color":"sky","label":"constant term","style":"solid"}]},"order":2,"title":"Apply the quadratic formula","description":"We identify the quadratic coefficients as $a = 1$, $b = -4$, and $c = 4 - y$, and substitute them into the quadratic formula $$x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$$"},{"type":"text","order":3,"title":"Simplify the expression","content":"Now, we simplify the terms inside and outside the square root:\n\n$$\\begin{aligned} x &= \\frac{4 \\pm \\sqrt{16 - 4(4 - y)}}{2} \\\\ x &= \\frac{4 \\pm \\sqrt{16 - 16 + 4y}}{2} \\\\ x &= \\frac{4 \\pm \\sqrt{4y}}{2} \\\\ x &= \\frac{4 \\pm 2\\sqrt{y}}{2} \\\\ x &= 2 \\pm \\sqrt{y} \\end{aligned}$$"},{"type":"text","order":4,"title":"Choose the correct sign","content":"Since the domain of the original function $f(x)$ is $x \\geq 2$, we must choose the positive sign to ensure our values of $x$ remain greater than or equal to $2$:\n\n$$x = 2 + \\sqrt{y}$$","highlights":[{"text":"x \\geq 2","location":"specification"}]},{"type":"text","order":5,"title":"Write the inverse function","content":"We now swap the variables $x$ and $y$ to obtain the final expression for the inverse function:\n\n$$f^{-1}(x) = \\sqrt{x} + 2$$"},{"type":"text","order":6,"title":"Determine the domain","content":"The domain of the inverse function $f^{-1}(x)$ is identical to the range of the original function $f(x)$.\n\nSince the minimum value of $f(x) = (x-2)^2$ occurs at $x=2$ where $f(2) = 0$, and $f(x)$ increases without bound for $x \\ge 2$, the range is $f(x) \\ge 0$.\n\nThus, the domain of $f^{-1}(x)$ is:\n\n$$x \\geq 0$$"}]}],"generatedAt":"2026-06-11T03:08:51.474Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"7e1f1c7a-7cee-4695-af29-4a10fd64bc2e:algebraic","label":"(a) Algebraic Method","steps":[{"type":"text","order":0,"title":"Understand the goal","content":"To find the range of the function $f(x)$, we need to determine the set of all possible output values (y-values) it can produce, given the domain constraint $x \\geq 2$. We will achieve this by rewriting the quadratic expression as a perfect square.","highlights":[{"text":"State the range of f.","location":"specification"},{"text":"f(x) = x^2 - 4x + 4, \\quad x \\in \\mathbb{R}, x \\geq 2","location":"specification"}]},{"type":"text","order":1,"title":"Factor into perfect square","content":"First, we factor the quadratic expression $x^2 - 4x + 4$ into a perfect square:\n\n$$f(x) = (x - 2)^2$$\n\nThis step earns the first method mark (**M1**)."},{"type":"text","order":2,"title":"Apply the domain constraint","content":"Now we use the given domain constraint, $x \\geq 2$, to find the possible values of $f(x)$:\n\n1. Start with the domain constraint:\n $$x \\geq 2$$\n2. Subtract $2$ from both sides:\n $$x - 2 \\geq 0$$\n3. Square both sides (since the terms are non-negative, the inequality direction is preserved):\n $$(x - 2)^2 \\geq 0$$"},{"type":"text","order":3,"title":"State the final range","content":"Since $f(x) = (x - 2)^2$ and $(x - 2)^2 \\geq 0$ for all $x \\geq 2$, the minimum value of $f(x)$ is $0$ (which occurs when $x = 2$). As $x$ increases, $(x-2)^2$ increases without limit.\n\nTherefore, the range of $f$ is:\n\n$$f(x) \\geq 0$$\n\nThis completes the second accuracy mark (**A1**)."}]},{"id":"7e1f1c7a-7cee-4695-af29-4a10fd64bc2e:graphical","label":"(a) Graphical Approach","steps":[{"type":"text","order":0,"title":"Analyze the function","content":"To find the range of $f(x) = x^2 - 4x + 4$ for $x \\geq 2$, we can analyze its graph. First, let's write the quadratic expression in vertex form by completing the square or factoring it.\n\nObserve that the quadratic is a perfect square trinomial:\n$$f(x) = (x - 2)^2$$\n\nThis form allows us to easily identify the key features of the parabola.","highlights":[{"text":"f(x) = x^2 - 4x + 4, \\quad x \\in \\mathbb{R}, x \\geq 2","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(x) = (x - \\htmlData{anchor=vertex-x}{\\color{@sky}{2}})^2 + \\htmlData{anchor=vertex-y}{\\color{@amber}{0}}$"},{"latex":"$$\\text{Vertex: } (\\htmlData{anchor=v-x}{\\color{@sky}{2}}, \\htmlData{anchor=v-y}{\\color{@amber}{0}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"vertex-x"},{"color":"amber","style":"text-color","anchor":"vertex-y"}],"connections":[{"to":"v-x","from":"vertex-x","color":"sky","label":"x-coord","style":"solid"},{"to":"v-y","from":"vertex-y","color":"amber","label":"y-coord","style":"solid"}]},"order":1,"title":"Identify the vertex","description":"From the vertex form, the vertex of the parabola is located at $(2, 0)$."},{"type":"text","order":2,"title":"Determine the shape and domain","content":"Let's consider the shape and the restricted domain of the function:\n- Since the coefficient of the $x^2$ term is positive ($+1$), the parabola opens upwards.\n- The domain of the function is restricted to $x \n\\geq 2$.\n\nBecause the domain starts exactly at the $x$-coordinate of the vertex ($x = 2$) and goes to positive infinity ($x \\to \\infty$), we are only looking at the right-hand branch of the parabola.","highlights":[{"text":"x \\geq 2","location":"specification"}]},{"type":"text","order":3,"title":"Find the range","content":"Since the graph starts at the vertex $(2, 0)$ and curves upwards as $x$ increases, the minimum value of $f(x)$ is the $y$-value of the vertex, which is $0$.\n\nAs $x$ increases towards infinity, $f(x)$ also increases towards infinity.\n\nTherefore, the range of the function is all values greater than or equal to $0$:\n$$f(x) \\geq 0 \\quad \\text{or} \\quad [0, \\infty)$$\n\nWriting this down as $f(x) \\geq 0$ (or $y \\geq 0$) will earn you full marks."}]},{"id":"b1cba5e7-cfe1-4c8b-9931-b27baf952724:geometric-reflection","label":"(b) Geometric Reflection","steps":[{"type":"text","order":0,"title":"Analyze the function","content":"First, let's write the given function in its completed square form to easily identify its shape and key points:\n\n$$f(x) = x^2 - 4x + 4 = (x-2)^2$$"},{"type":"text","order":1,"title":"Plot key points of f(x)","content":"Since the domain of $f(x)$ is $x \\ge 2$, we can determine a few key points on its curve:\n\n* **Starting point (vertex):** When $x = 2$, $f(2) = (2-2)^2 = 0$, giving the point $$(2, 0)$$\n* **Second point:** When $x = 3$, $f(3) = (3-2)^2 = 1$, giving the point $$(3, 1)$$\n* **Third point:** When $x = 4$, $f(4) = (4-2)^2 = 4$, giving the point $$(4, 4)$$\n\nDraw a smooth curve starting at $(2,0)$ and passing upwards through $(3,1)$ and $(4,4)$ to sketch $f(x)$."},{"type":"hyper_latex","block":{"nodes":[{"text":"An inverse function is a geometric reflection of the original function across the line $y=x$. This means we swap the $x$ and $y$ coordinates of each key point:"},{"latex":"$$\\htmlData{anchor=pt1}{(2, 0)} \\quad \\xrightarrow{\\text{reflect across } y=x} \\quad \\htmlData{anchor=pt1-ref}{(0, 2)}$"},{"latex":"$$\\htmlData{anchor=pt2}{(3, 1)} \\quad \\xrightarrow{\\text{reflect across } y=x} \\quad \\htmlData{anchor=pt2-ref}{(1, 3)}$"},{"latex":"$$\\htmlData{anchor=pt3}{(4, 4)} \\quad \\xrightarrow{\\text{reflect across } y=x} \\quad \\htmlData{anchor=pt3-ref}{(4, 4)}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"pt1"},{"color":"sky","style":"text-color","anchor":"pt1-ref"},{"color":"pink","style":"text-color","anchor":"pt2"},{"color":"pink","style":"text-color","anchor":"pt2-ref"}],"connections":[{"to":"pt1-ref","from":"pt1","color":"sky","label":"swap x and y","style":"solid"},{"to":"pt2-ref","from":"pt2","color":"pink","label":"swap x and y","style":"solid"}]},"order":2,"title":"Reflect points across y=x","description":"Each point $(a,b)$ on the graph of $f(x)$ corresponds to a point $(b,a)$ on the graph of $f^{-1}(x)$."},{"type":"text","order":3,"title":"Sketch both graphs","content":"To complete the sketch on your axes:\n\n1. Draw the line of symmetry $$y=x$$ as a dashed diagonal line.\n2. Plot the reflected points $(0, 2)$, $(1, 3)$, and $(4, 4)$.\n3. Draw a smooth curve starting from $(0, 2)$ on the $y$-axis and curving through $(1, 3)$ and $(4, 4)$ to represent $f^{-1}(x)$.\n\n### Markscheme Breakdown\n* **B1:** Curve for $f(x)$ starts at $(2,0)$ and passes through $(3,1)$.\n* **B1:** Curve for $f^{-1}(x)$ starts at $(0,2)$ and passes through $(1,3)$.\n* **B1:** The line of reflection $y=x$ is clearly shown, with both curves symmetric about it."}]},{"id":"b1cba5e7-cfe1-4c8b-9931-b27baf952724:algebraic-inverse","label":"(b) Algebraic Derivation and Plotting","steps":[{"type":"text","order":0,"title":"Analyze the function f(x)","content":"To sketch the graphs, we first examine the original function and its key features:\n\n$$f(x) = x^2 - 4x + 4, \\quad x \\geq 2$$\n\nThis can be rewritten by completing the square or factoring as a perfect square:\n\n$$f(x) = (x - 2)^2$$\n\nSince the domain is restricted to $x \\geq 2$, let's find two key coordinate points to plot this curve:\n- **Starting point**: When $x = 2$, $f(2) = (2 - 2)^2 = 0$. This gives the point $(2, 0)$.\n- **Another point**: When $x = 3$, $f(3) = (3 - 2)^2 = 1$. This gives the point $(3, 1)$.\n\nThe graph of $f(x)$ starts at $(2, 0)$ and curves upwards as a parabola through $(3, 1)$.","highlights":[{"text":"f(x) = x^2 - 4x + 4, \\quad x \\in \\mathbb{R}, x \\geq 2","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=f-eq}{(x - 2)^2}$"},{"latex":"$$\\sqrt{y} = \\htmlData{anchor=f-sqrt}{x - 2} \\quad \\text{since } x \\geq 2$"},{"latex":"$$x = \\htmlData{anchor=inv-y}{\\sqrt{y} + 2}$"},{"latex":"$$f^{-1}(x) = \\htmlData{anchor=inv-final}{\\sqrt{x} + 2}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"f-eq"},{"color":"amber","style":"box","anchor":"inv-final"}],"connections":[{"to":"f-sqrt","from":"f-eq","color":"sky","label":"square root","style":"solid"},{"to":"inv-final","from":"inv-y","color":"amber","label":"swap x and y","style":"solid"}]},"order":1,"title":"Find the inverse function","description":"We set $y = f(x)$ and solve for $x$ to find the algebraic equation of the inverse function $f^{-1}(x)$."},{"type":"text","order":2,"title":"Analyze the inverse function","content":"For the inverse function $f^{-1}(x) = \\sqrt{x} + 2$, the domain is the range of $f(x)$. Since $(x-2)^2 \\geq 0$, the domain of $f^{-1}(x)$ is $x \\geq 0$.\n\nLet's find key coordinate points for $f^{-1}(x)$:\n- **Starting point**: When $x = 0$, $f^{-1}(0) = \\sqrt{0} + 2 = 2$. This gives the point $(0, 2)$.\n- **Another point**: When $x = 1$, $f^{-1}(1) = \\sqrt{1} + 2 = 3$. This gives the point $(1, 3)$.\n\nNotice that these points are the exact reflections of the points of $f(x)$ across the line of symmetry $y = x$:\n- $(2, 0) \\leftrightarrow (0, 2)$\n- $(3, 1) \\leftrightarrow (1, 3)$"},{"type":"text","order":3,"title":"Sketch the curves","content":"Now, sketch both curves on the same Cartesian coordinate system:\n\n1. Draw the line of reflection $y = x$ as a dashed line.\n2. Plot and label $f(x)$ starting at $(2, 0)$, curving up through $(3, 1)$.\n3. Plot and label $f^{-1}(x)$ starting at $(0, 2)$, curving to the right through $(1, 3)$.\n4. Ensure the curves are symmetrical to each other across the line $y=x$.\n\n**Markscheme Tips:**\n- You get **1 mark** for correctly sketching $f(x)$ from $(2, 0)$ passing through $(3, 1)$.\n- You get **1 mark** for correctly sketching $f^{-1}(x)$ from $(0, 2)$ passing through $(1, 3)$.\n- You get **1 mark** for showing the line of symmetry $y=x$.","calculator":[{"gdc":{"instructions":"1. Go to Graph mode.\n2. Define \\\\f1(x) = (x-2)^2\\\\ with domain constraint \\\\x \\ge 2\\\\.\n3. Define \\\\f2(x) = \\sqrt{x} + 2\\\\.\n4. Define \\\\f3(x) = x\\\\.\n5. Set the window aspect ratio to square to visualize symmetry."},"ti84":{"instructions":"1. Press [Y=].\n2. Enter \\\\Y_1 = (X - 2)^2 / (X \\ge 2)\\\\ (use the logic test menu [2nd][MATH] for the inequality).\n3. Enter \\\\Y_2 = \\sqrt{X} + 2\\\\.\n4. Enter \\\\Y_3 = X\\\\.\n5. Press [ZOOM] [5] (ZSquare) to ensure equal scaling on both axes to see the reflection symmetry properly."},"desmos":{"instructions":"1. In the first input bar, type: \\\\f(x) = (x-2)^2 \\{x \\ge 2\\\\}.\n2. In the second input bar, type: \\\\g(x) = \\sqrt{x} + 2\\\\.\n3. In the third input bar, type: \\\\y = x\\\\ (change style to dashed in the settings)."},"description":"Using a graphing utility helps you visualize the curves and verify their symmetry across $y = x$."}]}]},{"id":"2161411d-0a4c-470d-901d-2f5e661b087c:completing_square","label":"(c) Completing the Square","steps":[{"type":"text","order":0,"title":"Complete the square","content":"To find the inverse function, we first express $f(x)$ in a completed square form.\n\nLooking at the quadratic expression:\n$$f(x) = x^2 - 4x + 4$$\n\nThis is a perfect square trinomial, which factorises directly to:\n$$y = (x - 2)^2$$\n\nWriting this as $y = (x - 2)^2$ earns you the first mark **(M1)**.","highlights":[{"text":"f(x) = x^2 - 4x + 4, \\quad x \\in \\mathbb{R}, x \\geq 2","location":"specification"}]},{"type":"text","order":1,"title":"Swap variables","content":"To find the inverse function, we swap the variables $x$ and $y$:\n$$x = (y - 2)^2$$"},{"type":"text","order":2,"title":"Solve for y","content":"Now we rearrange to solve for $y$ by taking the square root of both sides:\n$$\\sqrt{x} = \\pm(y - 2)$$\n\nSince the domain of our original function is restricted to $x \\geq 2$, the range of our inverse function $f^{-1}(x)$ (which corresponds to $y$) must also be $y \\geq 2$. Therefore, we must choose the positive square root:\n$$\\sqrt{x} = y - 2$$\n\nRearranging to make $y$ the subject:\n$$y = \\sqrt{x} + 2$$\n\nThis rearrangement and selection of the positive root earns you the second method mark **(M1)**.","highlights":[{"text":"x \\geq 2","location":"specification"}]},{"type":"text","order":3,"title":"State the inverse function","content":"We replace $y$ with $f^{-1}(x)$ to get our final inverse expression:\n$$f^{-1}(x) = \\sqrt{x} + 2$$\n\nThis correct expression earns the accuracy mark **(A1)**."},{"type":"text","order":4,"title":"Determine the domain","content":"The domain of the inverse function $f^{-1}(x)$ is equal to the range of the original function $f(x)$.\n\nGiven the domain of $f(x)$ is $x \\geq 2$, we can find the range of $f(x)$ by substituting the boundary value:\n- When $x = 2$, $f(2) = (2 - 2)^2 = 0$\n- For all $x \\geq 2$, $f(x) \\geq 0$\n\nTherefore, the range of $f(x)$ is $f(x) \\geq 0$.\n\nThis means the domain of $f^{-1}(x)$ is:\n$$x \\geq 0$$\n\nStating this domain correctly earns the final mark **(B1)**."}]},{"id":"2161411d-0a4c-470d-901d-2f5e661b087c:quadratic_formula","label":"(c) Quadratic Formula","steps":[{"type":"text","order":0,"title":"Set equal to y","content":"To find the inverse function, we first set the function expression equal to $y$:\n\n$$y = x^2 - 4x + 4$$"},{"type":"text","order":1,"title":"Rearrange to quadratic form","content":"Next, we rearrange the equation into a standard quadratic form $ax^2 + bx + c = 0$ with respect to $x$ by subtracting $y$ from both sides:\n\n$$x^2 - 4x + (4 - y) = 0$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2 - 4x + (\\htmlData{anchor=c-val}{\\color{@sky}{4 - y}}) = 0$"},{"latex":"$$x = \\frac{-(-4) \\pm \\sqrt{(-4)^2 - 4(1)(\\htmlData{anchor=c-sub}{\\color{@sky}{4 - y}})}}{2(1)}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"c-val"},{"color":"sky","style":"text-color","anchor":"c-sub"}],"connections":[{"to":"c-sub","from":"c-val","color":"sky","label":"constant term","style":"solid"}]},"order":2,"title":"Apply the quadratic formula","description":"We identify the quadratic coefficients as $a = 1$, $b = -4$, and $c = 4 - y$, and substitute them into the quadratic formula $$x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$$"},{"type":"text","order":3,"title":"Simplify the expression","content":"Now, we simplify the terms inside and outside the square root:\n\n$$\\begin{aligned} x &= \\frac{4 \\pm \\sqrt{16 - 4(4 - y)}}{2} \\\\ x &= \\frac{4 \\pm \\sqrt{16 - 16 + 4y}}{2} \\\\ x &= \\frac{4 \\pm \\sqrt{4y}}{2} \\\\ x &= \\frac{4 \\pm 2\\sqrt{y}}{2} \\\\ x &= 2 \\pm \\sqrt{y} \\end{aligned}$$"},{"type":"text","order":4,"title":"Choose the correct sign","content":"Since the domain of the original function $f(x)$ is $x \\geq 2$, we must choose the positive sign to ensure our values of $x$ remain greater than or equal to $2$:\n\n$$x = 2 + \\sqrt{y}$$","highlights":[{"text":"x \\geq 2","location":"specification"}]},{"type":"text","order":5,"title":"Write the inverse function","content":"We now swap the variables $x$ and $y$ to obtain the final expression for the inverse function:\n\n$$f^{-1}(x) = \\sqrt{x} + 2$$"},{"type":"text","order":6,"title":"Determine the domain","content":"The domain of the inverse function $f^{-1}(x)$ is identical to the range of the original function $f(x)$.\n\nSince the minimum value of $f(x) = (x-2)^2$ occurs at $x=2$ where $f(2) = 0$, and $f(x)$ increases without bound for $x \\ge 2$, the range is $f(x) \\ge 0$.\n\nThus, the domain of $f^{-1}(x)$ is:\n\n$$x \\geq 0$$"}]}],"generatedAt":"2026-06-11T03:08:51.969Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"b159a47b-6f72-407b-ba1c-c37aee3e6987","revisionId":"3d5d9ecc-6eff-47d2-8834-6481e983f2fc","questionNumber":null,"questionType":"LA","specification":"$$$f(x) = \\frac{2x+5}{x-3} \\quad x \\geq 5$$","difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"64de7522-0c1a-40c1-a256-c6f987784e9a","label":"a","content":"Find $ff(5)$","markscheme":"- $f(5) = \\frac{2(5)+5}{5-3} = \\frac{15}{2}$ M1\n- $f(\\frac{15}{2}) = \\frac{2(15/2)+5}{(15/2)-3} = \\frac{20}{9/2} = \\frac{40}{9}$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Evaluate $f(5)$ correctly to get $7.5$ or $\\frac{15}{2}$.","code":"M"},{"id":"p2","marks":1,"description":"Substitute their $f(5)$ back into the function and evaluate to $\\frac{40}{9}$.","code":"A"}],"finalAnswers":["\\frac{40}{9}","4.44"],"generatedAt":"2026-05-06T14:05:01.347Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":null,"options":[]},{"id":"76030b6c-8ae9-4aeb-b270-d321f22ab496","label":"b","content":"State the range of $f$.","markscheme":"- $f(5) = 7.5$. As $x \\to \\infty, f(x) \\to 2$, so $2 < f(x) \\leq 7.5$ B1","marks":1,"order":1,"markingRubric":{"totalMarks":1,"points":[{"id":"p1","marks":1,"description":"Correct range using appropriate inequality notation.","code":"B"}],"finalAnswers":["2 < f(x) \\leq 7.5"],"generatedAt":"2026-05-06T14:05:01.347Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":null,"options":[]},{"id":"b643ef91-a5c2-449c-b570-b3d8c0f61aff","label":"c","content":"Find $f^{-1}(x)$ and state its domain","markscheme":"- $y = \\frac{2x+5}{x-3} \\Rightarrow yx - 3y = 2x + 5$ M1\n- $x(y-2) = 5 + 3y \\Rightarrow x = \\frac{5+3y}{y-2}$ M1\n- $f^{-1}(x) = \\frac{3x+5}{x-2}$ with domain $2 < x \\leq 7.5$ A1","marks":3,"order":2,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Start process to find inverse by setting $y=f(x)$ and clearing the fraction.","code":"M"},{"id":"p2","marks":1,"description":"Isolate $x$ in terms of $y$.","code":"M"},{"id":"p3","marks":1,"description":"Correct inverse function expression and domain matching part (b).","code":"A"}],"finalAnswers":["f^{-1}(x) = \\frac{3x+5}{x-2}, 2 < x \\leq 7.5"],"generatedAt":"2026-05-06T14:05:01.347Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"7a5e88aa-a3c3-4fae-a5a7-be10bb775309","revisionId":"6ed8d273-a5be-43e5-bf29-efb1c0121b9d","questionNumber":null,"questionType":"LA","specification":"The function $f$ is defined by\n\n$$f(x) = x^2 + 8x + 16, \\quad x \\in \\mathbb{R}, x \\geq -4$$","difficultyBand":"medium","tier":null,"totalMarks":9,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"905c10db-26c0-4339-bc82-117605873e6f","label":"a","content":"State the range of $f$.","markscheme":"- $f(x) = (x+4)^2$ M1\n- $f(x) \\geq 0$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":1,"points":[{"id":"p1","marks":1,"description":"State $f(x) \\geq 0$.","code":"B"}],"finalAnswers":["$$f(x) \\geq 0$","$$y \\geq 0$"],"generatedAt":"2026-05-06T14:04:56.326Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"algebraic","label":"Algebraic Factoring","steps":[{"type":"text","order":0,"title":"Factor the quadratic expression","content":"To find the range of the function, we can start by factoring the quadratic. Let's rewrite the quadratic expression $f(x) = x^2 + 8x + 16$ as a perfect square.\n\nWe look for two numbers that multiply to $16$ and add to $8$. These numbers are $4$ and $4$:\n\n$$f(x) = (x+4)^2$$\n\nWriting the function in this form helps us easily identify how its values behave based on the input $x$.\n\n*Award 1 mark (M1) for writing the expression in its factored form.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Domain restriction: } \\htmlData{anchor=dom}{\\color{@sky}{x \\geq -4}}$"},{"latex":"$$\\htmlData{anchor=added}{\\color{@sky}{x + 4}} \\geq 0$"},{"latex":"$$f(x) = \\htmlData{anchor=sq}{\\color{@sky}{(x + 4)^2}} \\geq 0$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"dom"},{"color":"sky","style":"text-color","anchor":"added"},{"color":"sky","style":"text-color","anchor":"sq"}],"connections":[{"to":"added","from":"dom","color":"sky","label":"add 4 to both sides","style":"solid"},{"to":"sq","from":"added","color":"sky","label":"square both sides","style":"solid"}]},"order":1,"title":"Apply the domain restriction","description":"Now, we apply the given domain constraint to find the corresponding output values."},{"type":"text","order":2,"title":"State the range","content":"Since $(x+4)^2$ is always greater than or equal to $0$ for all $x \\geq -4$, the minimum value of $f(x)$ is $0$ (which occurs when $x = -4$). As $x$ increases, $(x+4)^2$ grows without bound.\n\nTherefore, the range of $f(x)$ is:\n\n$$f(x) \\geq 0 \\quad \\text{or} \\quad y \\geq 0$$\n\n*Award 1 mark (A1) for the correct range.*"}]},{"id":"graphical","label":"Graphical Analysis","steps":[{"type":"text","order":0,"title":"Analyze the function structure","content":"To understand the behavior of the function $f(x) = x^2 + 8x + 16$, we can rewrite it by factoring. \n\nRecognizing the quadratic as a perfect square trinomial:\n$$f(x) = (x + 4)^2$$ \n\nThis form lets us easily identify the shape and main features of its graph."},{"type":"text","order":1,"title":"Visualize the graph","content":"The graph of $y = (x + 4)^2$ is a translation of the basic parabola $y = x^2$ by $4$ units to the left.\n\nKey features of this parabola include:\n- It opens **upwards** because the coefficient of the squared term is positive ($1 > 0$).\n- Its minimum point (vertex) is located at $(-4, 0)$ where the squared term is at its minimum value of $0$."},{"type":"text","order":2,"title":"Apply the domain constraint","content":"The function is only defined for the restricted domain:\n$$x \\geq -4$$\n\nLooking at our graph starting from the vertex at $x = -4$ and moving to the right ($x \\geq -4$):\n- At the starting point $x = -4$, the function value is $f(-4) = (-4+4)^2 = 0$.\n- As $x$ increases from $-4$ towards positive infinity ($x \\to \\infty$), the curve rises continuously and $f(x)$ increases without bound."},{"type":"text","order":3,"title":"State the range","content":"Since the graph starts at a minimum $y$-value of $0$ when $x = -4$ and goes upwards indefinitely, the set of all possible output values (the range) is all real numbers greater than or equal to $0$.\n\nThus, the range is:\n$$f(x) \\geq 0 \\quad \\text{or} \\quad [0, \\infty)$$\n\n**Markscheme Guidance:**\n- **M1** is awarded for expressing the function as a perfect square, $f(x) = (x+4)^2$.\n- **A1** is awarded for identifying the correct range: $f(x) \\geq 0$."}]},{"id":"calculus","label":"Calculus Approach","steps":[{"type":"text","order":0,"title":"Differentiate the function","content":"To find the minimum value of the function, we can start by finding its first derivative. This allows us to locate any stationary points.\n\nGiven the function:\n$$f(x) = x^2 + 8x + 16$$\n\nWe differentiate with respect to $x$ using the power rule:\n$$f'(x) = 2x + 8$$"},{"type":"text","order":1,"title":"Locate the stationary point","content":"A stationary point occurs where the first derivative is equal to zero:\n$$f'(x) = 0$$\n\nSetting our derivative to $0$, we solve for $x$:\n$$\\begin{aligned} 2x + 8 &= 0 \\\\ 2x &= -8 \\\\ x &= -4 \\end{aligned}$$\n\nNotice that $x = -4$ is exactly the boundary of our domain, which is given as $x \\geq -4$."},{"type":"text","order":2,"title":"Calculate the minimum value","content":"Since the coefficient of $x^2$ is positive, the stationary point at $x = -4$ is a local (and global) minimum. \n\nWe calculate the minimum value of the function by substituting $x = -4$ back into the original expression for $f(x)$:\n$$\\begin{aligned} f(-4) &= (-4)^2 + 8(-4) + 16 \\\\ f(-4) &= 16 - 32 + 16 \\\\ f(-4) &= 0 \\end{aligned}$$"},{"type":"text","order":3,"title":"Determine the range","content":"Because the function has a minimum value of $0$ at $x = -4$ and increases continuously as $x$ increases beyond $-4$, the output values of the function will always be greater than or equal to $0$.\n\nTherefore, the range of $f$ is:\n$$f(x) \\geq 0$$","calculator":[{"gdc":{"expression":"x^2 + 8x + 16","instructions":"1. Go to the Graph menu.\n2. Enter the function y1 = x^2 + 8x + 16.\n3. Draw the graph and use the G-Solv (or Analyze Graph) tool.\n4. Select 'MIN' to find the minimum point, which will display as (-4, 0)."},"ti84":{"expression":"x^2 + 8x + 16","instructions":"1. Press [Y=] and enter Y1 = X^2 + 8X + 16.\n2. Press [WINDOW] and set Xmin = -5, Xmax = 5, Ymin = -2, Ymax = 20.\n3. Press [GRAPH] to view the curve.\n4. Press [2nd] [TRACE] (CALC) and select '3: minimum'.\n5. Set a Left Bound to the left of x = -4, a Right Bound to the right, and press [ENTER] to find the minimum point at (-4, 0)."},"desmos":{"expression":"y = x^2 + 8x + 16 \\\\left\\\\{x \\\\ge -4\\\\right\\\\}","instructions":"1. Type y = x^2 + 8x + 16 {x >= -4} into the first expression line.\n2. Click on the curve to highlight key points.\n3. Click on the gray dot at the lowest point of the curve to reveal the coordinates (-4, 0)."},"description":"You can verify the minimum point and the range by graphing the function on your calculator."}]}]}],"generatedAt":"2026-06-09T06:03:36.450Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"c75b8360-3f24-422f-b254-61dd1b8ee4eb","label":"b","content":"Sketch the graphs of $f(x)$ and $f^{-1}(x)$ on the same diagram.","markscheme":"- Sketch of $f(x)$ starting at $(-4, 0)$ and passing through $(0, 16)$ B1\n- Sketch of $f^{-1}(x)$ starting at $(0, -4)$ and passing through $(16, 0)$ B1\n- Reflection in the line $y=x$ shown B1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Correct shape for $f(x)$ starting at $(-4, 0)$.","code":"B"},{"id":"p2","marks":1,"description":"Correct shape for $f^{-1}(x)$ starting at $(0, -4)$.","code":"B"},{"id":"p3","marks":1,"description":"Line $y=x$ shown as a line of symmetry.","code":"B"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:04:56.326Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"geometric-reflection","label":"Geometric Reflection Method","steps":[{"type":"text","order":0,"title":"Analyze the function f(x)","content":"The given function is:\n\n$$f(x) = x^2 + 8x + 16, \\quad x \\geq -4$$\n\nFirst, let's rewrite it in vertex form by factoring the perfect square trinomial:\n\n$$f(x) = (x+4)^2$$\n\nFrom this form, we can identify two key features of the graph of $f(x)$:\n- **Vertex**: The vertex of the parabola is at $(-4, 0)$. Since the domain is restricted to $x \\geq -4$, this vertex is the starting point of our curve.\n- **$y$-intercept**: When $x = 0$, $f(0) = (0+4)^2 = 16$, giving the point $(0, 16)$."},{"type":"text","order":1,"title":"Sketch the curve of f(x)","content":"Using the key features we found, we can sketch the graph of $f(x)$:\n\n1. Plot the starting endpoint at the vertex $(-4, 0)$.\n2. Plot the $y$-intercept at $(0, 16)$.\n3. Draw a smooth, upward-sloping half-parabola starting from $(-4,0)$ and passing through $(0, 16)$ into the first quadrant."},{"type":"text","order":2,"title":"Reflect across the mirror line","content":"Because $f^{-1}(x)$ is the inverse of $f(x)$, its graph is a reflection of the graph of $f(x)$ across the diagonal line $y = x$.\n\nTo find the coordinates of the key points on $f^{-1}(x)$, we simply swap the $x$ and $y$ coordinates of the key points on $f(x)$:\n- The starting point $(-4, 0)$ reflects to $(0, -4)$.\n- The intercept $(0, 16)$ reflects to $(16, 0)$."},{"type":"text","order":3,"title":"Complete the joint sketch","content":"Now, draw both curves and the line of reflection on the same set of axes:\n\n1. Draw the dashed line $y = x$ passing through the origin.\n2. Sketch the curve $f^{-1}(x)$ starting at $(0, -4)$ on the $y$-axis, curving up to pass through $(16, 0)$ on the $x$-axis.\n3. Make sure the curves of $f(x)$ and $f^{-1}(x)$ look symmetric with respect to the line $y=x$.\n\n### Markscheme Points:\n- **B1**: Correct sketch of $f(x)$ starting at $(-4, 0)$ and passing through $(0, 16)$.\n- **B1**: Correct sketch of $f^{-1}(x)$ starting at $(0, -4)$ and passing through $(16, 0)$.\n- **B1**: The line of reflection $y=x$ is drawn and labelled."}]},{"id":"algebraic-inverse","label":"Analytical Inverse Method","steps":[{"type":"text","order":0,"title":"Complete the square","content":"To find the analytical expression for the inverse function $f^{-1}(x)$, we first rewrite $f(x)$ by completing the square.\n\n$$f(x) = x^2 + 8x + 16$$\n\nWe recognize this quadratic expression as a perfect square trinomial:\n\n$$f(x) = (x + 4)^2$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=eq-form}{(x + 4)^2}$"},{"latex":"$$\\pm\\sqrt{y} = x + 4$"},{"latex":"$$\\text{Since } x \\geq -4 \\text{, we have } x + 4 \\geq 0 \\implies \\htmlData{anchor=pos-root}{\\sqrt{y}} = x + 4$"},{"latex":"$$x = \\htmlData{anchor=inv-expr}{\\sqrt{y} - 4}$"}],"highlights":[{"color":"sky","style":"box","anchor":"eq-form"},{"color":"amber","style":"text-color","anchor":"inv-expr"}],"connections":[{"to":"pos-root","from":"eq-form","color":"sky","label":"take square root","style":"solid"},{"to":"inv-expr","from":"pos-root","color":"amber","label":"solve for x","style":"solid"}]},"order":1,"title":"Find the inverse function","description":"We set $y = f(x)$ and rearrange to make $x$ the subject. Because of the domain restriction $x \\geq -4$, we choose the positive square root."},{"type":"text","order":2,"title":"Define the inverse","content":"By swapping $x$ and $y$, we obtain the algebraic equation for the inverse function:\n\n$$f^{-1}(x) = \\sqrt{x} - 4$$\n\nThe domain of $f^{-1}(x)$ is the range of $f(x)$. Since $f(x) = (x+4)^2 \\geq 0$ for $x \\geq -4$, the domain of $f^{-1}(x)$ is $x \\geq 0$."},{"type":"text","order":3,"title":"Determine key sketch points","content":"To sketch both curves accurately, we identify key points on each graph:\n\n- **For $f(x) = (x+4)^2$ where $x \\geq -4$**:\n - The starting point (vertex) is at $(-4, 0)$.\n - The $y$-intercept is at $(0, 16)$ (since $f(0) = 16$).\n\n- **For $f^{-1}(x) = \\sqrt{x} - 4$ where $x \\geq 0$**:\n - The starting point is at $(0, -4)$.\n - The $x$-intercept is at $(16, 0)$ (since $f^{-1}(16) = \\sqrt{16} - 4 = 0$).\n\nNotice that the coordinates are swapped, which confirms they are reflections of each other."},{"type":"text","order":4,"title":"Sketch the graphs","content":"Now, sketch both curves on the same axes showing their symmetry:\n\n1. Draw the line of symmetry $$y = x$$ as a dashed diagonal line.\n2. Plot $f(x)$ starting at $(-4, 0)$, curving upwards through the point $(0, 16)$. This represents one branch of a parabola.\n3. Plot $f^{-1}(x)$ starting at $(0, -4)$, curving to the right through the point $(16, 0)$. This represents a square-root curve.\n4. Ensure the curves look like mirror images of each other across the line $y = x$.\n\n### Markscheme Breakdown:\n- **B1**: Sketching $f(x)$ with correct start point $(-4, 0)$ and passing through $(0, 16)$.\n- **B1**: Sketching $f^{-1}(x)$ with correct start point $(0, -4)$ and passing through $(16, 0)$.\n- **B1**: Showing clear graphical symmetry / reflection in the line $y=x$."}]}],"generatedAt":"2026-06-09T06:05:16.829Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"f7b808ab-8648-42e3-ab81-ff9c850888dd","label":"c","content":"Find an expression for $f^{-1}(x)$ and state its domain.","markscheme":"- $y = (x+4)^2 \\Rightarrow \\sqrt{y} = x+4$ M1\n- $x = \\sqrt{y} - 4$ M1\n- $f^{-1}(x) = \\sqrt{x} - 4$ A1\n- Domain: $x \\geq 0$ B1","marks":4,"order":2,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Factorises to $(x+4)^2$ and attempts to isolate $x$.","code":"M"},{"id":"p2","marks":1,"description":"Correct algebraic rearrangement to $x = \\sqrt{y} - 4$.","code":"M"},{"id":"p3","marks":1,"description":"Correct inverse function $f^{-1}(x) = \\sqrt{x} - 4$.","code":"A"},{"id":"p4","marks":1,"description":"Correct domain for the inverse: $x \\geq 0$.","code":"B"}],"finalAnswers":["$$f^{-1}(x) = \\sqrt{x} - 4, x \\geq 0$"],"generatedAt":"2026-05-06T14:04:56.326Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"perfect_square_factorization","label":"Perfect Square Factorization","steps":[{"type":"text","order":0,"title":"Recognize the perfect square","content":"Let's set $y = f(x)$ and observe the quadratic expression:\n\n$$y = x^2 + 8x + 16$$\n\nWe can recognize this expression as a perfect square trinomial because it fits the pattern $a^2 + 2ab + b^2 = (a+b)^2$, where $a=x$ and $b=4$:\n\n$$y = (x + 4)^2$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\left(\\htmlData{anchor=x-expr}{x + 4}\\right)^2$"},{"latex":"$$\\pm\\htmlData{anchor=sqrt-y}{\\sqrt{y}} = x + 4$"}],"highlights":[{"color":"sky","style":"box","anchor":"x-expr"},{"color":"sky","style":"box","anchor":"sqrt-y"}],"connections":[{"to":"sqrt-y","from":"x-expr","color":"sky","label":"take square root","style":"solid"}]},"order":1,"title":"Rearrange for x","description":"To make $x$ the subject, we take the square root of both sides. This gives us two possible branches: positive or negative square root."},{"type":"text","order":2,"title":"Determine the correct sign","content":"We are given the domain constraint for the original function:\n\n$$x \\geq -4 \\implies x + 4 \\geq 0$$\n\nSince $x + 4$ must be non-negative, we must choose the positive square root:\n\n$$\\sqrt{y} = x + 4$$\n\nNow, isolate $x$ by subtracting $4$ from both sides:\n\n$$x = \\sqrt{y} - 4$$"},{"type":"text","order":3,"title":"Write the inverse function","content":"To write the final expression for the inverse function $f^{-1}(x)$, we swap the variables $x$ and $y$:\n\n$$f^{-1}(x) = \\sqrt{x} - 4$$"},{"type":"text","order":4,"title":"Determine the domain","content":"The domain of the inverse function $f^{-1}(x)$ is equal to the range of the original function $f(x)$. Since the domain of $f(x)$ is $x \\geq -4$:\n\n$$x + 4 \\geq 0 \\implies y = (x + 4)^2 \\geq 0$$\n\nTherefore, the range of $f(x)$ is $f(x) \\geq 0$, which means the domain of $f^{-1}(x)$ is:\n\n$$x \\geq 0$$"}]},{"id":"quadratic_formula","label":"Quadratic Formula Approach","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"To find the inverse function, we start by setting the function equal to $y$:\n\n$$y = x^2 + 8x + 16$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2 + 8x + \\htmlData{anchor=const}{\\color{@sky}{16 - y}} = 0$"}],"highlights":[{"color":"sky","style":"box","anchor":"const"}],"connections":[]},"order":1,"title":"Rearrange to standard form","description":"We rearrange the equation into a standard quadratic equation in terms of $x$, of the form $ax^2 + bx + c = 0$, where $a = 1$, $b = 8$, and the constant term $c = 16 - y$."},{"type":"text","order":2,"title":"Apply the quadratic formula","content":"We can solve for $x$ using the quadratic formula with $a = 1$, $b = 8$, and $c = 16 - y$. Let's first evaluate the discriminant term under the square root to keep things simple:\n\n$$b^2 - 4ac = 8^2 - 4(1)(16 - y)$$\n\n$$b^2 - 4ac = 64 - 64 + 4y = 4y$$"},{"type":"text","order":3,"title":"Simplify the expression","content":"Now substitute the simplified discriminant back into the quadratic formula:\n\n$$x = \\frac{-8 \\pm \\sqrt{4y}}{2(1)}$$\n\nSince $\\sqrt{4y} = 2\\sqrt{y}$, this simplifies directly to:\n\n$$x = \\frac{-8 \\pm 2\\sqrt{y}}{2} = -4 \\pm \\sqrt{y}$$"},{"type":"text","order":4,"title":"Choose the correct root","content":"Since the domain of the original function $f(x)$ is defined as $x \\geq -4$, we must choose the positive sign option to ensure $x$ is greater than or equal to $-4$:\n\n$$x = -4 + \\sqrt{y}$$"},{"type":"text","order":5,"title":"Determine the domain","content":"The domain of the inverse function $f^{-1}(x)$ is equal to the range of the original function $f(x)$ for its defined domain $x \\geq -4$. Since $f(x) = (x+4)^2$, we can evaluate the function at its boundary $x = -4$:\n\n$$f(-4) = (-4 + 4)^2 = 0$$\n\nSince $x \\ge -4$, the value of $(x+4)^2$ will always be greater than or equal to $0$. Thus, the range of $f$ is $y \\geq 0$. Therefore, the domain of $f^{-1}(x)$ is:\n\n$$x \\geq 0$$"},{"type":"text","order":6,"title":"Write final inverse function","content":"We replace $y$ with $x$ to write the expression for the inverse function:\n\n$$f^{-1}(x) = \\sqrt{x} - 4 \\quad \\text{for } x \\geq 0$$"}]}],"generatedAt":"2026-06-09T06:07:31.616Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"905c10db-26c0-4339-bc82-117605873e6f:algebraic","label":"(a) Algebraic Factoring","steps":[{"type":"text","order":0,"title":"Factor the quadratic expression","content":"To find the range of the function, we can start by factoring the quadratic. Let's rewrite the quadratic expression $f(x) = x^2 + 8x + 16$ as a perfect square.\n\nWe look for two numbers that multiply to $16$ and add to $8$. These numbers are $4$ and $4$:\n\n$$f(x) = (x+4)^2$$\n\nWriting the function in this form helps us easily identify how its values behave based on the input $x$.\n\n*Award 1 mark (M1) for writing the expression in its factored form.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Domain restriction: } \\htmlData{anchor=dom}{\\color{@sky}{x \\geq -4}}$"},{"latex":"$$\\htmlData{anchor=added}{\\color{@sky}{x + 4}} \\geq 0$"},{"latex":"$$f(x) = \\htmlData{anchor=sq}{\\color{@sky}{(x + 4)^2}} \\geq 0$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"dom"},{"color":"sky","style":"text-color","anchor":"added"},{"color":"sky","style":"text-color","anchor":"sq"}],"connections":[{"to":"added","from":"dom","color":"sky","label":"add 4 to both sides","style":"solid"},{"to":"sq","from":"added","color":"sky","label":"square both sides","style":"solid"}]},"order":1,"title":"Apply the domain restriction","description":"Now, we apply the given domain constraint to find the corresponding output values."},{"type":"text","order":2,"title":"State the range","content":"Since $(x+4)^2$ is always greater than or equal to $0$ for all $x \\geq -4$, the minimum value of $f(x)$ is $0$ (which occurs when $x = -4$). As $x$ increases, $(x+4)^2$ grows without bound.\n\nTherefore, the range of $f(x)$ is:\n\n$$f(x) \\geq 0 \\quad \\text{or} \\quad y \\geq 0$$\n\n*Award 1 mark (A1) for the correct range.*"}]},{"id":"905c10db-26c0-4339-bc82-117605873e6f:graphical","label":"(a) Graphical Analysis","steps":[{"type":"text","order":0,"title":"Analyze the function structure","content":"To understand the behavior of the function $f(x) = x^2 + 8x + 16$, we can rewrite it by factoring. \n\nRecognizing the quadratic as a perfect square trinomial:\n$$f(x) = (x + 4)^2$$ \n\nThis form lets us easily identify the shape and main features of its graph."},{"type":"text","order":1,"title":"Visualize the graph","content":"The graph of $y = (x + 4)^2$ is a translation of the basic parabola $y = x^2$ by $4$ units to the left.\n\nKey features of this parabola include:\n- It opens **upwards** because the coefficient of the squared term is positive ($1 > 0$).\n- Its minimum point (vertex) is located at $(-4, 0)$ where the squared term is at its minimum value of $0$."},{"type":"text","order":2,"title":"Apply the domain constraint","content":"The function is only defined for the restricted domain:\n$$x \\geq -4$$\n\nLooking at our graph starting from the vertex at $x = -4$ and moving to the right ($x \\geq -4$):\n- At the starting point $x = -4$, the function value is $f(-4) = (-4+4)^2 = 0$.\n- As $x$ increases from $-4$ towards positive infinity ($x \\to \\infty$), the curve rises continuously and $f(x)$ increases without bound."},{"type":"text","order":3,"title":"State the range","content":"Since the graph starts at a minimum $y$-value of $0$ when $x = -4$ and goes upwards indefinitely, the set of all possible output values (the range) is all real numbers greater than or equal to $0$.\n\nThus, the range is:\n$$f(x) \\geq 0 \\quad \\text{or} \\quad [0, \\infty)$$\n\n**Markscheme Guidance:**\n- **M1** is awarded for expressing the function as a perfect square, $f(x) = (x+4)^2$.\n- **A1** is awarded for identifying the correct range: $f(x) \\geq 0$."}]},{"id":"905c10db-26c0-4339-bc82-117605873e6f:calculus","label":"(a) Calculus Approach","steps":[{"type":"text","order":0,"title":"Differentiate the function","content":"To find the minimum value of the function, we can start by finding its first derivative. This allows us to locate any stationary points.\n\nGiven the function:\n$$f(x) = x^2 + 8x + 16$$\n\nWe differentiate with respect to $x$ using the power rule:\n$$f'(x) = 2x + 8$$"},{"type":"text","order":1,"title":"Locate the stationary point","content":"A stationary point occurs where the first derivative is equal to zero:\n$$f'(x) = 0$$\n\nSetting our derivative to $0$, we solve for $x$:\n$$\\begin{aligned} 2x + 8 &= 0 \\\\ 2x &= -8 \\\\ x &= -4 \\end{aligned}$$\n\nNotice that $x = -4$ is exactly the boundary of our domain, which is given as $x \\geq -4$."},{"type":"text","order":2,"title":"Calculate the minimum value","content":"Since the coefficient of $x^2$ is positive, the stationary point at $x = -4$ is a local (and global) minimum. \n\nWe calculate the minimum value of the function by substituting $x = -4$ back into the original expression for $f(x)$:\n$$\\begin{aligned} f(-4) &= (-4)^2 + 8(-4) + 16 \\\\ f(-4) &= 16 - 32 + 16 \\\\ f(-4) &= 0 \\end{aligned}$$"},{"type":"text","order":3,"title":"Determine the range","content":"Because the function has a minimum value of $0$ at $x = -4$ and increases continuously as $x$ increases beyond $-4$, the output values of the function will always be greater than or equal to $0$.\n\nTherefore, the range of $f$ is:\n$$f(x) \\geq 0$$","calculator":[{"gdc":{"expression":"x^2 + 8x + 16","instructions":"1. Go to the Graph menu.\n2. Enter the function y1 = x^2 + 8x + 16.\n3. Draw the graph and use the G-Solv (or Analyze Graph) tool.\n4. Select 'MIN' to find the minimum point, which will display as (-4, 0)."},"ti84":{"expression":"x^2 + 8x + 16","instructions":"1. Press [Y=] and enter Y1 = X^2 + 8X + 16.\n2. Press [WINDOW] and set Xmin = -5, Xmax = 5, Ymin = -2, Ymax = 20.\n3. Press [GRAPH] to view the curve.\n4. Press [2nd] [TRACE] (CALC) and select '3: minimum'.\n5. Set a Left Bound to the left of x = -4, a Right Bound to the right, and press [ENTER] to find the minimum point at (-4, 0)."},"desmos":{"expression":"y = x^2 + 8x + 16 \\\\left\\\\{x \\\\ge -4\\\\right\\\\}","instructions":"1. Type y = x^2 + 8x + 16 {x >= -4} into the first expression line.\n2. Click on the curve to highlight key points.\n3. Click on the gray dot at the lowest point of the curve to reveal the coordinates (-4, 0)."},"description":"You can verify the minimum point and the range by graphing the function on your calculator."}]}]},{"id":"c75b8360-3f24-422f-b254-61dd1b8ee4eb:geometric-reflection","label":"(b) Geometric Reflection Method","steps":[{"type":"text","order":0,"title":"Analyze the function f(x)","content":"The given function is:\n\n$$f(x) = x^2 + 8x + 16, \\quad x \\geq -4$$\n\nFirst, let's rewrite it in vertex form by factoring the perfect square trinomial:\n\n$$f(x) = (x+4)^2$$\n\nFrom this form, we can identify two key features of the graph of $f(x)$:\n- **Vertex**: The vertex of the parabola is at $(-4, 0)$. Since the domain is restricted to $x \\geq -4$, this vertex is the starting point of our curve.\n- **$y$-intercept**: When $x = 0$, $f(0) = (0+4)^2 = 16$, giving the point $(0, 16)$."},{"type":"text","order":1,"title":"Sketch the curve of f(x)","content":"Using the key features we found, we can sketch the graph of $f(x)$:\n\n1. Plot the starting endpoint at the vertex $(-4, 0)$.\n2. Plot the $y$-intercept at $(0, 16)$.\n3. Draw a smooth, upward-sloping half-parabola starting from $(-4,0)$ and passing through $(0, 16)$ into the first quadrant."},{"type":"text","order":2,"title":"Reflect across the mirror line","content":"Because $f^{-1}(x)$ is the inverse of $f(x)$, its graph is a reflection of the graph of $f(x)$ across the diagonal line $y = x$.\n\nTo find the coordinates of the key points on $f^{-1}(x)$, we simply swap the $x$ and $y$ coordinates of the key points on $f(x)$:\n- The starting point $(-4, 0)$ reflects to $(0, -4)$.\n- The intercept $(0, 16)$ reflects to $(16, 0)$."},{"type":"text","order":3,"title":"Complete the joint sketch","content":"Now, draw both curves and the line of reflection on the same set of axes:\n\n1. Draw the dashed line $y = x$ passing through the origin.\n2. Sketch the curve $f^{-1}(x)$ starting at $(0, -4)$ on the $y$-axis, curving up to pass through $(16, 0)$ on the $x$-axis.\n3. Make sure the curves of $f(x)$ and $f^{-1}(x)$ look symmetric with respect to the line $y=x$.\n\n### Markscheme Points:\n- **B1**: Correct sketch of $f(x)$ starting at $(-4, 0)$ and passing through $(0, 16)$.\n- **B1**: Correct sketch of $f^{-1}(x)$ starting at $(0, -4)$ and passing through $(16, 0)$.\n- **B1**: The line of reflection $y=x$ is drawn and labelled."}]},{"id":"c75b8360-3f24-422f-b254-61dd1b8ee4eb:algebraic-inverse","label":"(b) Analytical Inverse Method","steps":[{"type":"text","order":0,"title":"Complete the square","content":"To find the analytical expression for the inverse function $f^{-1}(x)$, we first rewrite $f(x)$ by completing the square.\n\n$$f(x) = x^2 + 8x + 16$$\n\nWe recognize this quadratic expression as a perfect square trinomial:\n\n$$f(x) = (x + 4)^2$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=eq-form}{(x + 4)^2}$"},{"latex":"$$\\pm\\sqrt{y} = x + 4$"},{"latex":"$$\\text{Since } x \\geq -4 \\text{, we have } x + 4 \\geq 0 \\implies \\htmlData{anchor=pos-root}{\\sqrt{y}} = x + 4$"},{"latex":"$$x = \\htmlData{anchor=inv-expr}{\\sqrt{y} - 4}$"}],"highlights":[{"color":"sky","style":"box","anchor":"eq-form"},{"color":"amber","style":"text-color","anchor":"inv-expr"}],"connections":[{"to":"pos-root","from":"eq-form","color":"sky","label":"take square root","style":"solid"},{"to":"inv-expr","from":"pos-root","color":"amber","label":"solve for x","style":"solid"}]},"order":1,"title":"Find the inverse function","description":"We set $y = f(x)$ and rearrange to make $x$ the subject. Because of the domain restriction $x \\geq -4$, we choose the positive square root."},{"type":"text","order":2,"title":"Define the inverse","content":"By swapping $x$ and $y$, we obtain the algebraic equation for the inverse function:\n\n$$f^{-1}(x) = \\sqrt{x} - 4$$\n\nThe domain of $f^{-1}(x)$ is the range of $f(x)$. Since $f(x) = (x+4)^2 \\geq 0$ for $x \\geq -4$, the domain of $f^{-1}(x)$ is $x \\geq 0$."},{"type":"text","order":3,"title":"Determine key sketch points","content":"To sketch both curves accurately, we identify key points on each graph:\n\n- **For $f(x) = (x+4)^2$ where $x \\geq -4$**:\n - The starting point (vertex) is at $(-4, 0)$.\n - The $y$-intercept is at $(0, 16)$ (since $f(0) = 16$).\n\n- **For $f^{-1}(x) = \\sqrt{x} - 4$ where $x \\geq 0$**:\n - The starting point is at $(0, -4)$.\n - The $x$-intercept is at $(16, 0)$ (since $f^{-1}(16) = \\sqrt{16} - 4 = 0$).\n\nNotice that the coordinates are swapped, which confirms they are reflections of each other."},{"type":"text","order":4,"title":"Sketch the graphs","content":"Now, sketch both curves on the same axes showing their symmetry:\n\n1. Draw the line of symmetry $$y = x$$ as a dashed diagonal line.\n2. Plot $f(x)$ starting at $(-4, 0)$, curving upwards through the point $(0, 16)$. This represents one branch of a parabola.\n3. Plot $f^{-1}(x)$ starting at $(0, -4)$, curving to the right through the point $(16, 0)$. This represents a square-root curve.\n4. Ensure the curves look like mirror images of each other across the line $y = x$.\n\n### Markscheme Breakdown:\n- **B1**: Sketching $f(x)$ with correct start point $(-4, 0)$ and passing through $(0, 16)$.\n- **B1**: Sketching $f^{-1}(x)$ with correct start point $(0, -4)$ and passing through $(16, 0)$.\n- **B1**: Showing clear graphical symmetry / reflection in the line $y=x$."}]},{"id":"f7b808ab-8648-42e3-ab81-ff9c850888dd:perfect_square_factorization","label":"(c) Perfect Square Factorization","steps":[{"type":"text","order":0,"title":"Recognize the perfect square","content":"Let's set $y = f(x)$ and observe the quadratic expression:\n\n$$y = x^2 + 8x + 16$$\n\nWe can recognize this expression as a perfect square trinomial because it fits the pattern $a^2 + 2ab + b^2 = (a+b)^2$, where $a=x$ and $b=4$:\n\n$$y = (x + 4)^2$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\left(\\htmlData{anchor=x-expr}{x + 4}\\right)^2$"},{"latex":"$$\\pm\\htmlData{anchor=sqrt-y}{\\sqrt{y}} = x + 4$"}],"highlights":[{"color":"sky","style":"box","anchor":"x-expr"},{"color":"sky","style":"box","anchor":"sqrt-y"}],"connections":[{"to":"sqrt-y","from":"x-expr","color":"sky","label":"take square root","style":"solid"}]},"order":1,"title":"Rearrange for x","description":"To make $x$ the subject, we take the square root of both sides. This gives us two possible branches: positive or negative square root."},{"type":"text","order":2,"title":"Determine the correct sign","content":"We are given the domain constraint for the original function:\n\n$$x \\geq -4 \\implies x + 4 \\geq 0$$\n\nSince $x + 4$ must be non-negative, we must choose the positive square root:\n\n$$\\sqrt{y} = x + 4$$\n\nNow, isolate $x$ by subtracting $4$ from both sides:\n\n$$x = \\sqrt{y} - 4$$"},{"type":"text","order":3,"title":"Write the inverse function","content":"To write the final expression for the inverse function $f^{-1}(x)$, we swap the variables $x$ and $y$:\n\n$$f^{-1}(x) = \\sqrt{x} - 4$$"},{"type":"text","order":4,"title":"Determine the domain","content":"The domain of the inverse function $f^{-1}(x)$ is equal to the range of the original function $f(x)$. Since the domain of $f(x)$ is $x \\geq -4$:\n\n$$x + 4 \\geq 0 \\implies y = (x + 4)^2 \\geq 0$$\n\nTherefore, the range of $f(x)$ is $f(x) \\geq 0$, which means the domain of $f^{-1}(x)$ is:\n\n$$x \\geq 0$$"}]},{"id":"f7b808ab-8648-42e3-ab81-ff9c850888dd:quadratic_formula","label":"(c) Quadratic Formula Approach","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"To find the inverse function, we start by setting the function equal to $y$:\n\n$$y = x^2 + 8x + 16$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2 + 8x + \\htmlData{anchor=const}{\\color{@sky}{16 - y}} = 0$"}],"highlights":[{"color":"sky","style":"box","anchor":"const"}],"connections":[]},"order":1,"title":"Rearrange to standard form","description":"We rearrange the equation into a standard quadratic equation in terms of $x$, of the form $ax^2 + bx + c = 0$, where $a = 1$, $b = 8$, and the constant term $c = 16 - y$."},{"type":"text","order":2,"title":"Apply the quadratic formula","content":"We can solve for $x$ using the quadratic formula with $a = 1$, $b = 8$, and $c = 16 - y$. Let's first evaluate the discriminant term under the square root to keep things simple:\n\n$$b^2 - 4ac = 8^2 - 4(1)(16 - y)$$\n\n$$b^2 - 4ac = 64 - 64 + 4y = 4y$$"},{"type":"text","order":3,"title":"Simplify the expression","content":"Now substitute the simplified discriminant back into the quadratic formula:\n\n$$x = \\frac{-8 \\pm \\sqrt{4y}}{2(1)}$$\n\nSince $\\sqrt{4y} = 2\\sqrt{y}$, this simplifies directly to:\n\n$$x = \\frac{-8 \\pm 2\\sqrt{y}}{2} = -4 \\pm \\sqrt{y}$$"},{"type":"text","order":4,"title":"Choose the correct root","content":"Since the domain of the original function $f(x)$ is defined as $x \\geq -4$, we must choose the positive sign option to ensure $x$ is greater than or equal to $-4$:\n\n$$x = -4 + \\sqrt{y}$$"},{"type":"text","order":5,"title":"Determine the domain","content":"The domain of the inverse function $f^{-1}(x)$ is equal to the range of the original function $f(x)$ for its defined domain $x \\geq -4$. Since $f(x) = (x+4)^2$, we can evaluate the function at its boundary $x = -4$:\n\n$$f(-4) = (-4 + 4)^2 = 0$$\n\nSince $x \\ge -4$, the value of $(x+4)^2$ will always be greater than or equal to $0$. Thus, the range of $f$ is $y \\geq 0$. Therefore, the domain of $f^{-1}(x)$ is:\n\n$$x \\geq 0$$"},{"type":"text","order":6,"title":"Write final inverse function","content":"We replace $y$ with $x$ to write the expression for the inverse function:\n\n$$f^{-1}(x) = \\sqrt{x} - 4 \\quad \\text{for } x \\geq 0$$"}]}],"generatedAt":"2026-06-09T06:07:44.943Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths","summaries":[{"id":"777f03b9-6dd4-46eb-a3e8-0ec448bdc187","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"7be6b99e-7a00-4d93-ac55-e63bc9f644ee","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"ab41448b-8e3d-4c93-a90f-88f19401fb2f","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"b159a47b-6f72-407b-ba1c-c37aee3e6987","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"7a5e88aa-a3c3-4fae-a5a7-be10bb775309","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"a4b9c690-7281-4267-b63d-55b8d4b207a0","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"f8374545-57a2-42f9-afd6-97b4e373f677","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"3f00a3c9-1db7-42f3-9f45-673b29a8b5b3","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"bf853a17-d712-49c0-b3c9-f0b8ac478631","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"c6eab48b-b642-4ff8-9908-9091145e2a8a","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"022bbde6-3afb-4e73-93a7-1040bf21753c","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"0a74abb2-d4db-4253-98f6-6a922128d703","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"105daad0-1fa5-43c3-b75a-185e2223392b","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"1c2390d9-8c1c-41cd-a98e-f0cfc28f2b14","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"246022b6-a8e7-4c4c-acd2-9d07583c4910","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"2f6e853d-78f4-4e50-8842-1fdb947bed20","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"462900d5-bc5c-4c15-b953-c2b0e2811138","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"4ca481df-f794-4d3d-a310-73b98456e3c9","difficultyBand":"medium","calculatorUse":null,"totalMarks":11,"questionType":"LA","isCalculation":true},{"id":"57396252-3ae9-4ae8-901a-9ce64c6c9d53","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"5ca0480b-b65e-49be-8de9-4b0e9dc66d46","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"602feb89-78c5-4d29-80a5-b3a1f179a090","difficultyBand":"easy","calculatorUse":null,"totalMarks":1,"questionType":"LA","isCalculation":true},{"id":"6337b580-ff93-4a0a-b82d-fb84b8bfd81b","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"74b1d1c4-0239-4469-a696-fed801817642","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"7d78ab6c-d1ff-4193-b01b-e27a0980ac87","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"8a758b82-bf4e-4be9-8d8e-045d5a6ff039","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"907c00a1-dcea-4ce0-9d2b-6ad8f59488f8","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"90a54c95-25fe-457d-bb8d-92ab6872beb7","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"a2a43ef4-7897-4b4a-b97c-0eb6d63c2814","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"a84544f2-3213-4dd4-8344-146417ff4216","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"aa0a64d3-d981-4152-a758-3d6be2cc2e5b","difficultyBand":"hard","calculatorUse":null,"totalMarks":13,"questionType":"LA","isCalculation":true},{"id":"c7e3f284-a124-4040-ab0d-2933e73672c0","difficultyBand":"hard","calculatorUse":null,"totalMarks":12,"questionType":"LA","isCalculation":true},{"id":"cd5129b1-8362-4035-89db-3b09f4b08e8f","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"d0b1d97c-89a4-4350-b82f-f0a8a24b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Inverse Functions","topicId":"2425a502-b65e-4ac9-99ad-3700e78c1711","topicSidebar":"$L3b","topicSlug":"a-level-pure-mathematics-2-4-inverse-functions","topicTitle":"2.4 Inverse Functions"}]}],"$L3c","$L3d"]}]

2.4 Inverse Functions

Question 3

2.4 Inverse Functions Questions

2.4 Inverse Functions

Question 3

2.4 Inverse Functions Questions