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A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Attempt to complete the square, identifying $(x-5)^2$ or equivalent method to find the vertex.","code":"M"},{"id":"p2","marks":1,"description":"Correctly identifying the minimum value of the function as 5.","code":"M"},{"id":"p3","marks":1,"description":"Correct range stated with appropriate notation.","code":"A","acceptable":["f(x) >= 5","y >= 5","[5, \\infty)"]}],"finalAnswers":["$$f(x) \\geq 5$"],"generatedAt":"2026-05-06T14:04:48.764Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"completing_the_square","label":"Completing the Square","steps":[{"type":"text","order":0,"title":"Understand the Goal","content":"To find the range of the quadratic function $f(x) = x^2 - 10x + 30$, we need to find the set of all possible outputs $f(x)$ for real inputs $x$. Since the coefficient of the $x^2$ term is positive ($1 > 0$), the graph of this function is a parabola that opens upwards. This means it has a minimum value but no maximum value. \n\nWe will use the **Completing the Square** method to rewrite the function in its vertex form, $f(x) = (x-h)^2 + k$, which will allow us to immediately identify the minimum value."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$f(x) = \\htmlData{anchor=orig}{x^2 - 10x} + 30$$"},{"latex":"$$$f(x) = \\htmlData{anchor=square}{(x - 5)^2 - 25} + 30$$"},{"latex":"$$$f(x) = (x - 5)^2 + \\htmlData{anchor=final-k}{\\color{@amber}{5}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"orig"},{"color":"sky","style":"text-color","anchor":"square"},{"color":"amber","style":"box","anchor":"final-k"}],"connections":[{"to":"square","from":"orig","color":"sky","label":"Complete the square","style":"solid"}]},"order":1,"title":"Complete the Square","description":"To complete the square for $x^2 - 10x$, take half of the $x$-coefficient: $\\frac{-10}{2} = -5$. Squaring this gives $(-5)^2 = 25$. We write the first two terms as $(x-5)^2 - 25$, then add the constant $30$."},{"type":"text","order":2,"title":"Identify the Minimum Value","content":"Now we analyze our completed-square form:\n$$f(x) = (x-5)^2 + 5$$\n\nBecause $(x-5)^2$ is a squared term, it must be greater than or equal to zero for any real value of $x$:\n$$(x-5)^2 \\ge 0$$\n\nThus, the minimum possible value of $(x-5)^2$ is $0$, which occurs when $x = 5$. At this point, the function reaches its minimum value:\n$$f(5) = 0 + 5 = 5$$\n\nThis means the vertex (minimum point) of the quadratic curve is at $(5, 5)$."},{"type":"text","order":3,"title":"State the Range","content":"Since the function opens upwards and has a minimum value of $5$, the output values $f(x)$ can be any real number greater than or equal to $5$.\n\nTherefore, the range of $f(x)$ is:\n$$f(x) \\ge 5$$\n\n**Marks breakdown:**\n* **M1**: Correctly completing the square to get $(x-5)^2 + 5$.\n* **M1**: Identifying the minimum point of the function is at $y = 5$.\n* **A1**: Correctly writing the final range as $f(x) \\ge 5$."}]},{"id":"vertex_formula","label":"Using the Vertex Formula","steps":[{"type":"text","order":0,"title":"Identify coefficients","content":"The given function is a quadratic function:\n\n$$f(x) = x^2 - 10x + 30$$\n\nTo find its range, we first need to determine the minimum value of this quadratic curve. Since the coefficient of $x^2$ is positive ($a = 1 > 0$), the graph of $f(x)$ is a parabola that opens upwards, meaning the vertex represents the absolute minimum point of the function.\n\nLet us identify the coefficients of $f(x) = ax^2 + bx + c$:\n- $a = 1$\n- $b = -10$\n- $c = 30$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x = \\frac{-\\htmlData{anchor=b-val}{\\color{@red}{b}}}{2\\htmlData{anchor=a-val}{\\color{@sky}{a}}}$"},{"latex":"$$x = \\frac{-(\\htmlData{anchor=b-sub}{\\color{@red}{-10}})}{2(\\htmlData{anchor=a-sub}{\\color{@sky}{1}})}$"},{"latex":"$$x = \\frac{10}{2} = 5$"}],"highlights":[{"color":"red","style":"text-color","anchor":"b-val"},{"color":"sky","style":"text-color","anchor":"a-val"}],"connections":[{"to":"b-sub","from":"b-val","color":"red","label":null,"style":"solid"},{"to":"a-sub","from":"a-val","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Find the x-coordinate","description":"We can find the $x$-coordinate of the vertex using the vertex formula $x = -\\frac{b}{2a}$."},{"type":"text","order":2,"title":"Find the minimum value","content":"Now we substitute our $x$-coordinate back into the original function $f(x)$ to find the minimum value of the function:\n\n$$\\begin{aligned} f(5) &= (5)^2 - 10(5) + 30 \\\\ &= 25 - 50 + 30 \\\\ &= 5 \\end{aligned}$$\n\nThis gives us the coordinates of the vertex at $(5, 5)$, which corresponds to the minimum point on the curve."},{"type":"text","order":3,"title":"State the range","content":"Since the minimum value of the function is $5$ and the function goes up to infinity, the range must be all real values greater than or equal to $5$.\n\nTherefore, the range of $f(x)$ is:\n\n$$f(x) \\ge 5$$"}]},{"id":"differentiation","label":"Using Calculus (Differentiation)","steps":[{"type":"text","order":0,"title":"Understand the goal","content":"We want to find the range of the function:\n\n$$f(x) = x^2 - 10x + 30, \\quad x \\in \\mathbb{R}$$\n\nUsing calculus, we can find the minimum point of this curve. Since this is a quadratic curve with a positive $x^2$ coefficient (it opens upwards), the range will be all values of $f(x)$ greater than or equal to its minimum value."},{"type":"text","order":1,"title":"Find the derivative","content":"First, we find the first derivative of $f(x)$ with respect to $x$ using the power rule:\n\n$$f'(x) = \\frac{d}{dx}(x^2 - 10x + 30)$$\n\n$$f'(x) = 2x - 10$$"},{"type":"text","order":2,"title":"Locate the minimum","content":"To find the stationary point, we set the first derivative equal to $0$:\n\n$$\\begin{aligned} 2x - 10 &= 0 \\\\ 2x &= 10 \\\\ x &= 5 \\end{aligned}$$\n\nBecause the second derivative is positive ($f''(x) = 2 > 0$), this confirms that $x = 5$ is a local and global minimum point."},{"type":"text","order":3,"title":"Calculate the minimum value","content":"Now, we evaluate $f(x)$ at $x = 5$ to find the minimum $y$-coordinate of the curve:\n\n$$\\begin{aligned} f(5) &= (5)^2 - 10(5) + 30 \\\\ &= 25 - 50 + 30 \\\\ &= 5 \\end{aligned}$$\n\nThus, the minimum value of $f(x)$ is $5$."},{"type":"text","order":4,"title":"State the range","content":"Since the minimum value of the function is $5$ and the function increases without bound on either side, the range of $f(x)$ is:\n\n$$f(x) \\geq 5$$"}]}],"generatedAt":"2026-06-09T07:24:18.928Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"242fdce0-061a-4c81-8205-4d35e1792bd6","label":"b","content":"Explain why the function $f(x)$ does not have an inverse","markscheme":"- It is a many-to-one function (many-to-one functions do not have an inverse) B1","marks":1,"order":1,"markingRubric":{"totalMarks":1,"points":[{"id":"p1","marks":1,"description":"State that the function is many-to-one.","code":"B","acceptable":["It is many-to-one","Fails the horizontal line test","Not one-to-one"]}],"finalAnswers":[],"generatedAt":"2026-05-06T14:04:48.764Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"mapping_argument","label":"Using Function Mapping (Many-to-One)","steps":[{"type":"text","order":0,"title":"Recall inverse conditions","content":"For a function to have an inverse, it must be a **one-to-one** function.\n\nThis means that every distinct input $x$ in the domain must map to a unique output $y$ in the range. If a function is **many-to-one**, meaning multiple different inputs map to the same output, then an inverse function cannot exist because the inverse process would not yield a single unique output for that value."},{"type":"text","order":1,"title":"Find mapping counterexample","content":"We can show that $f(x) = x^2 - 10x + 30$ is many-to-one by finding two different $x$-values that produce the exact same output.\n\nLet's test two values that are symmetric around the line of symmetry of the quadratic curve, which is at $x = 5$:\n\nFor $x = 4$:\n$$f(4) = (4)^2 - 10(4) + 30 = 16 - 40 + 30 = 6$$\n\nFor $x = 6$:\n$$f(6) = (6)^2 - 10(6) + 30 = 36 - 60 + 30 = 6$$"},{"type":"text","order":2,"title":"Formulate conclusion","content":"Since $f(4) = f(6) = 6$, we have demonstrated that two distinct input values ($4 \\neq 6$) map to the same output value ($6$).\n\nTherefore, the function $f(x)$ is **many-to-one** and does not have an inverse.\n\n*(Note: To get the mark in the exam, simply stating that the function is \"many-to-one\" is sufficient!)*"}]},{"id":"graphical_horizontal_line","label":"Using the Horizontal Line Test","steps":[{"type":"text","order":0,"title":"Recall inverse conditions","content":"For a function to have an inverse, it must be a **one-to-one** function. This means that every output value $y$ must map back to exactly one unique input value $x$.\n\nIf a function is **many-to-one** (where more than one $x$-value yields the same $y$-value), its inverse does not exist because we wouldn't know which $x$ to return to."},{"type":"text","order":1,"title":"Analyze the function's graph","content":"The given function is:\n\n$$f(x) = x^2 - 10x + 30, \\quad x \\in \\mathbb{R}$$\n\nThis is a quadratic function, so its graph is a **parabola** that opens upwards. Because the domain is defined for all real numbers ($x \\in \\mathbb{R}$), the graph includes both the left and right symmetrical halves of the parabola on either side of its turning point (vertex)."},{"type":"text","order":2,"title":"Apply the Horizontal Line Test","content":"We can test if a function is one-to-one graphically using the **Horizontal Line Test**:\n\n- If any horizontal line $y = k$ intersects the graph at **more than one point**, the function is **many-to-one** and does not have an inverse.\n\nFor our parabola, its vertex is at $(5, 5)$. If we draw any horizontal line above this turning point (for example, the line $y = 6$), it will intersect the curve at two distinct points:\n\n$$x^2 - 10x + 30 = 6$$\n\n$$x^2 - 10x + 24 = 0$$\n\n$$(x - 4)(x - 6) = 0$$\n\nThis gives two inputs, $x = 4$ and $x = 6$, that yield the exact same output of $6$ ($f(4) = f(6) = 6$)."},{"type":"text","order":3,"title":"Conclude the explanation","content":"Since a horizontal line can intersect the graph of $f(x)$ at more than one point, the function fails the horizontal line test.\n\nTherefore, $f(x)$ is a **many-to-one** function and **does not have an inverse**."}]}],"generatedAt":"2026-06-09T07:26:35.508Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"7672515b-c9ad-40a1-893e-9bbd1aae90e7:completing_the_square","label":"(a) Completing the Square","steps":[{"type":"text","order":0,"title":"Understand the Goal","content":"To find the range of the quadratic function $f(x) = x^2 - 10x + 30$, we need to find the set of all possible outputs $f(x)$ for real inputs $x$. Since the coefficient of the $x^2$ term is positive ($1 > 0$), the graph of this function is a parabola that opens upwards. This means it has a minimum value but no maximum value. \n\nWe will use the **Completing the Square** method to rewrite the function in its vertex form, $f(x) = (x-h)^2 + k$, which will allow us to immediately identify the minimum value."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$f(x) = \\htmlData{anchor=orig}{x^2 - 10x} + 30$$"},{"latex":"$$$f(x) = \\htmlData{anchor=square}{(x - 5)^2 - 25} + 30$$"},{"latex":"$$$f(x) = (x - 5)^2 + \\htmlData{anchor=final-k}{\\color{@amber}{5}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"orig"},{"color":"sky","style":"text-color","anchor":"square"},{"color":"amber","style":"box","anchor":"final-k"}],"connections":[{"to":"square","from":"orig","color":"sky","label":"Complete the square","style":"solid"}]},"order":1,"title":"Complete the Square","description":"To complete the square for $x^2 - 10x$, take half of the $x$-coefficient: $\\frac{-10}{2} = -5$. Squaring this gives $(-5)^2 = 25$. We write the first two terms as $(x-5)^2 - 25$, then add the constant $30$."},{"type":"text","order":2,"title":"Identify the Minimum Value","content":"Now we analyze our completed-square form:\n$$f(x) = (x-5)^2 + 5$$\n\nBecause $(x-5)^2$ is a squared term, it must be greater than or equal to zero for any real value of $x$:\n$$(x-5)^2 \\ge 0$$\n\nThus, the minimum possible value of $(x-5)^2$ is $0$, which occurs when $x = 5$. At this point, the function reaches its minimum value:\n$$f(5) = 0 + 5 = 5$$\n\nThis means the vertex (minimum point) of the quadratic curve is at $(5, 5)$."},{"type":"text","order":3,"title":"State the Range","content":"Since the function opens upwards and has a minimum value of $5$, the output values $f(x)$ can be any real number greater than or equal to $5$.\n\nTherefore, the range of $f(x)$ is:\n$$f(x) \\ge 5$$\n\n**Marks breakdown:**\n* **M1**: Correctly completing the square to get $(x-5)^2 + 5$.\n* **M1**: Identifying the minimum point of the function is at $y = 5$.\n* **A1**: Correctly writing the final range as $f(x) \\ge 5$."}]},{"id":"7672515b-c9ad-40a1-893e-9bbd1aae90e7:vertex_formula","label":"(a) Using the Vertex Formula","steps":[{"type":"text","order":0,"title":"Identify coefficients","content":"The given function is a quadratic function:\n\n$$f(x) = x^2 - 10x + 30$$\n\nTo find its range, we first need to determine the minimum value of this quadratic curve. Since the coefficient of $x^2$ is positive ($a = 1 > 0$), the graph of $f(x)$ is a parabola that opens upwards, meaning the vertex represents the absolute minimum point of the function.\n\nLet us identify the coefficients of $f(x) = ax^2 + bx + c$:\n- $a = 1$\n- $b = -10$\n- $c = 30$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x = \\frac{-\\htmlData{anchor=b-val}{\\color{@red}{b}}}{2\\htmlData{anchor=a-val}{\\color{@sky}{a}}}$"},{"latex":"$$x = \\frac{-(\\htmlData{anchor=b-sub}{\\color{@red}{-10}})}{2(\\htmlData{anchor=a-sub}{\\color{@sky}{1}})}$"},{"latex":"$$x = \\frac{10}{2} = 5$"}],"highlights":[{"color":"red","style":"text-color","anchor":"b-val"},{"color":"sky","style":"text-color","anchor":"a-val"}],"connections":[{"to":"b-sub","from":"b-val","color":"red","label":null,"style":"solid"},{"to":"a-sub","from":"a-val","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Find the x-coordinate","description":"We can find the $x$-coordinate of the vertex using the vertex formula $x = -\\frac{b}{2a}$."},{"type":"text","order":2,"title":"Find the minimum value","content":"Now we substitute our $x$-coordinate back into the original function $f(x)$ to find the minimum value of the function:\n\n$$\\begin{aligned} f(5) &= (5)^2 - 10(5) + 30 \\\\ &= 25 - 50 + 30 \\\\ &= 5 \\end{aligned}$$\n\nThis gives us the coordinates of the vertex at $(5, 5)$, which corresponds to the minimum point on the curve."},{"type":"text","order":3,"title":"State the range","content":"Since the minimum value of the function is $5$ and the function goes up to infinity, the range must be all real values greater than or equal to $5$.\n\nTherefore, the range of $f(x)$ is:\n\n$$f(x) \\ge 5$$"}]},{"id":"7672515b-c9ad-40a1-893e-9bbd1aae90e7:differentiation","label":"(a) Using Calculus (Differentiation)","steps":[{"type":"text","order":0,"title":"Understand the goal","content":"We want to find the range of the function:\n\n$$f(x) = x^2 - 10x + 30, \\quad x \\in \\mathbb{R}$$\n\nUsing calculus, we can find the minimum point of this curve. Since this is a quadratic curve with a positive $x^2$ coefficient (it opens upwards), the range will be all values of $f(x)$ greater than or equal to its minimum value."},{"type":"text","order":1,"title":"Find the derivative","content":"First, we find the first derivative of $f(x)$ with respect to $x$ using the power rule:\n\n$$f'(x) = \\frac{d}{dx}(x^2 - 10x + 30)$$\n\n$$f'(x) = 2x - 10$$"},{"type":"text","order":2,"title":"Locate the minimum","content":"To find the stationary point, we set the first derivative equal to $0$:\n\n$$\\begin{aligned} 2x - 10 &= 0 \\\\ 2x &= 10 \\\\ x &= 5 \\end{aligned}$$\n\nBecause the second derivative is positive ($f''(x) = 2 > 0$), this confirms that $x = 5$ is a local and global minimum point."},{"type":"text","order":3,"title":"Calculate the minimum value","content":"Now, we evaluate $f(x)$ at $x = 5$ to find the minimum $y$-coordinate of the curve:\n\n$$\\begin{aligned} f(5) &= (5)^2 - 10(5) + 30 \\\\ &= 25 - 50 + 30 \\\\ &= 5 \\end{aligned}$$\n\nThus, the minimum value of $f(x)$ is $5$."},{"type":"text","order":4,"title":"State the range","content":"Since the minimum value of the function is $5$ and the function increases without bound on either side, the range of $f(x)$ is:\n\n$$f(x) \\geq 5$$"}]},{"id":"242fdce0-061a-4c81-8205-4d35e1792bd6:mapping_argument","label":"(b) Using Function Mapping (Many-to-One)","steps":[{"type":"text","order":0,"title":"Recall inverse conditions","content":"For a function to have an inverse, it must be a **one-to-one** function.\n\nThis means that every distinct input $x$ in the domain must map to a unique output $y$ in the range. If a function is **many-to-one**, meaning multiple different inputs map to the same output, then an inverse function cannot exist because the inverse process would not yield a single unique output for that value."},{"type":"text","order":1,"title":"Find mapping counterexample","content":"We can show that $f(x) = x^2 - 10x + 30$ is many-to-one by finding two different $x$-values that produce the exact same output.\n\nLet's test two values that are symmetric around the line of symmetry of the quadratic curve, which is at $x = 5$:\n\nFor $x = 4$:\n$$f(4) = (4)^2 - 10(4) + 30 = 16 - 40 + 30 = 6$$\n\nFor $x = 6$:\n$$f(6) = (6)^2 - 10(6) + 30 = 36 - 60 + 30 = 6$$"},{"type":"text","order":2,"title":"Formulate conclusion","content":"Since $f(4) = f(6) = 6$, we have demonstrated that two distinct input values ($4 \\neq 6$) map to the same output value ($6$).\n\nTherefore, the function $f(x)$ is **many-to-one** and does not have an inverse.\n\n*(Note: To get the mark in the exam, simply stating that the function is \"many-to-one\" is sufficient!)*"}]},{"id":"242fdce0-061a-4c81-8205-4d35e1792bd6:graphical_horizontal_line","label":"(b) Using the Horizontal Line Test","steps":[{"type":"text","order":0,"title":"Recall inverse conditions","content":"For a function to have an inverse, it must be a **one-to-one** function. This means that every output value $y$ must map back to exactly one unique input value $x$.\n\nIf a function is **many-to-one** (where more than one $x$-value yields the same $y$-value), its inverse does not exist because we wouldn't know which $x$ to return to."},{"type":"text","order":1,"title":"Analyze the function's graph","content":"The given function is:\n\n$$f(x) = x^2 - 10x + 30, \\quad x \\in \\mathbb{R}$$\n\nThis is a quadratic function, so its graph is a **parabola** that opens upwards. Because the domain is defined for all real numbers ($x \\in \\mathbb{R}$), the graph includes both the left and right symmetrical halves of the parabola on either side of its turning point (vertex)."},{"type":"text","order":2,"title":"Apply the Horizontal Line Test","content":"We can test if a function is one-to-one graphically using the **Horizontal Line Test**:\n\n- If any horizontal line $y = k$ intersects the graph at **more than one point**, the function is **many-to-one** and does not have an inverse.\n\nFor our parabola, its vertex is at $(5, 5)$. If we draw any horizontal line above this turning point (for example, the line $y = 6$), it will intersect the curve at two distinct points:\n\n$$x^2 - 10x + 30 = 6$$\n\n$$x^2 - 10x + 24 = 0$$\n\n$$(x - 4)(x - 6) = 0$$\n\nThis gives two inputs, $x = 4$ and $x = 6$, that yield the exact same output of $6$ ($f(4) = f(6) = 6$)."},{"type":"text","order":3,"title":"Conclude the explanation","content":"Since a horizontal line can intersect the graph of $f(x)$ at more than one point, the function fails the horizontal line test.\n\nTherefore, $f(x)$ is a **many-to-one** function and **does not have an inverse**."}]}],"generatedAt":"2026-06-09T07:26:45.323Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":47},"walkthroughVideo":null},{"id":"5d3e02dc-cec3-4cac-9d59-24032bd4142d","revisionId":"8f1687bb-8abd-4059-a999-250c1b72a388","questionNumber":null,"questionType":"LA","specification":"The function $g$ is defined by\n\n$$g(x) = x^2 + 6x - 2, \\quad x \\in \\mathbb{R}$$","difficulty":1,"difficultyBand":"medium","totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"11a3f5e5-6f30-4b6a-989a-cc97f78da5c8","label":"a","content":"Find the range of $g(x)$","markscheme":"- $g(x) = (x+3)^2 - 9 - 2 = (x+3)^2 - 11$ M1\n- Minimum point is $(-3, -11)$ M1\n- Range: $g(x) \\geq -11$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Attempt to complete the square resulting in $(x+3)^2$ or using $-b/2a$ to find the $x$-coordinate of the vertex.","code":"M"},{"id":"p2","marks":1,"description":"Finding the minimum value of $-11$.","code":"M"},{"id":"p3","marks":1,"description":"Correct range $g(x) \\geq -11$.","code":"A","acceptable":["g(x) >= -11","y >= -11"]}],"finalAnswers":["$$g(x) \\geq -11$"],"generatedAt":"2026-05-06T14:04:50.857Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"completing_the_square","label":"Completing the Square","steps":[{"type":"text","order":0,"title":"Understand the task","content":"To find the range of the function $g(x) = x^2 + 6x - 2$, we want to find all the possible output values of $g(x)$ for any real input $x$.\n\nSince this is a quadratic function with a positive $x^2$ coefficient, its graph is a u-shaped parabola. Its range will be all values greater than or equal to its minimum value.\n\nWe can find this minimum value by rewriting the quadratic in **vertex form**, $g(x) = (x-h)^2 + k$, using the **completing the square** method."},{"type":"text","order":1,"title":"Complete the square","content":"We focus on the first two terms, $x^2 + 6x$. We halve the coefficient of $x$ (which is $6$) to get $3$, and then subtract its square ($3^2 = 9$):\n\n$$\\begin{aligned} g(x) &= x^2 + 6x - 2 \\\\ &= (x + 3)^2 - 9 - 2 \\\\ &= (x + 3)^2 - 11 \\end{aligned}$$"},{"type":"text","order":2,"title":"Find the minimum value","content":"Now that we have $g(x) = (x+3)^2 - 11$, let's analyze the expression to find its minimum value:\n\n* The squared term, $(x+3)^2$, is always non-negative because any real number squared is greater than or equal to zero:\n $$(x+3)^2 \\ge 0$$\n* The minimum possible value of this squared term is $0$, which occurs when $x = -3$.\n* Substituting $(x+3)^2 = 0$ gives the minimum value of the function:\n $$g(-3) = 0 - 11 = -11$$\n\nThus, the minimum point (vertex) of the curve is $(-3, -11)$.."},{"type":"text","order":3,"title":"State the range","content":"Since the minimum value of $g(x)$ is $-11$, the function can take any value greater than or equal to $-11$.\n\nTherefore, the range of $g(x)$ is:\n\n$$g(x) \\ge -11$$\n\n### Markscheme Breakdown\n* **M1**: Correctly completing the square to get $g(x) = (x+3)^2 - 11$\n* **M1**: Identifying the minimum point as $(-3, -11)$\n* **A1**: Writing the range as $g(x) \\ge -11$ (or $y \\ge -11$)"},{"type":"text","order":4,"title":"Explain why no inverse exists","content":"For a function to have an inverse, it must be a **one-to-one** function (meaning every output $y$ corresponds to exactly one input $x$).\n\nSince $g(x) = x^2 + 6x - 2$ is a quadratic function, its graph is a parabola. For any value in its range (except the minimum value of $-11$), there are two different input values of $x$ that map to the same output $g(x)$.\n\nFor example, both $x = 0$ and $x = -6$ map to $-2$:\n$$g(0) = (0)^2 + 6(0) - 2 = -2$$\n$$g(-6) = (-6)^2 + 6(-6) - 2 = -2$$\n\nSince it is a **many-to-one** function rather than a one-to-one function, $g(x)$ does not have an inverse.\n\n### Markscheme Breakdown\n* **B1**: Explaining that $g(x)$ is a many-to-one function (or that two distinct $x$ values yield the same output value)."}]},{"id":"vertex_formula","label":"Using the Vertex Formula","steps":[{"type":"text","order":0,"title":"Identify the coefficients","content":"To find the vertex of the quadratic function $g(x) = x^2 + 6x - 2$, we first identify the coefficients from the general quadratic form $g(x) = ax^2 + bx + c$:\n\n* $$a = 1$$\n* $$b = 6$$\n* $$c = -2$$"},{"type":"text","order":1,"title":"Calculate vertex x-coordinate","content":"We use the vertex formula $x = -\\frac{b}{2a}$ to find the $x$-coordinate of the vertex of the parabola.\n\nSubstituting $a = 1$ and $b = 6$:\n\n$$x = -\\frac{6}{2(1)} = -3$$"},{"type":"text","order":2,"title":"Find the minimum value","content":"Now we evaluate $g(x)$ at $x = -3$ to find the corresponding $y$-coordinate of the vertex:\n\n$$\\begin{aligned} g(-3) &= (-3)^2 + 6(-3) - 2 \\\\ &= 9 - 18 - 2 \\\\ &= -11 \\end{aligned}$$\n\nSo, the vertex of the parabola is at $(-3, -11)$.."},{"type":"text","order":3,"title":"Determine the range","content":"Since the coefficient of $x^2$ is positive ($a = 1 > 0$), the parabola opens upwards.\n\nThis means the vertex represents the absolute minimum point of the function.\n\nTherefore, the range of $g(x)$ is all values greater than or equal to this minimum:\n\n$$g(x) \\geq -11$$","calculator":[{"gdc":{"expression":"x^2 + 6*x - 2","instructions":"1. Go to the Graph application.\n2. Enter the function \$f(x) = x^2 + 6x - 2\$.\n3. Use the 'Analyze Graph' or 'G-Solv' tool and select 'Minimum'.\n4. The calculator will display the vertex coordinates at \$(-3, -11)\$."},"ti84":{"expression":"X^2 + 6*X - 2","instructions":"1. Press [Y=] and enter \$Y_1 = X^2 + 6X - 2\$.\n2. Press [GRAPH] to view the curve.\n3. Press [2nd] [TRACE] (CALC) and select '3: minimum'.\n4. Set a 'Left Bound' to the left of the vertex, a 'Right Bound' to the right, and press [ENTER] on 'Guess'.\n5. The screen will display the coordinates of the minimum point at approximately \$x = -3, y = -11\$."},"desmos":{"expression":"y = x^2 + 6*x - 2","instructions":"1. Type `y = x^2 + 6x - 2` into the input bar.\n2. Click on the curve to reveal key points.\n3. Click on the gray dot at the very bottom of the parabola to see the coordinates of the minimum vertex point: \$(-3, -11)\$."},"description":"You can verify this range visually by graphing the function and finding its minimum value."}]},{"type":"text","order":4,"title":"Explain why no inverse exists","content":"For a function to have an inverse, it must be a **one-to-one** function (meaning every output $y$ corresponds to exactly one input $x$).\n\nSince $g(x) = x^2 + 6x - 2$ is a quadratic function, its graph is a parabola. For any value in its range (except the minimum value of $-11$), there are two different input values of $x$ that map to the same output $g(x)$.\n\nFor example, both $x = 0$ and $x = -6$ map to $-2$:\n$$g(0) = (0)^2 + 6(0) - 2 = -2$$\n$$g(-6) = (-6)^2 + 6(-6) - 2 = -2$$\n\nSince it is a **many-to-one** function rather than a one-to-one function, $g(x)$ does not have an inverse.\n\n### Markscheme Breakdown\n* **B1**: Explaining that $g(x)$ is a many-to-one function (or that two distinct $x$ values yield the same output value)."}]},{"id":"calculus","label":"Using Differentiation","steps":[{"type":"text","order":0,"title":"Understand the goal","content":"We want to find the range of the quadratic function:\n\n$$g(x) = x^2 + 6x - 2$$\n\nBecause the coefficient of $x^2$ is positive ($1 > 0$), the graph of this function is a $U$-shaped parabola that opens upwards. This means the function has a global minimum point. The range will include all values greater than or equal to this minimum value:\n\n$$\\text{Range: } g(x) \\geq g(x_{\\text{min}})$$\n\nWe will use differentiation to locate this minimum point."},{"type":"text","order":1,"title":"Find the derivative","content":"Differentiate $g(x)$ term-by-term with respect to $x$ using the power rule $\\frac{d}{dx}(x^n) = n x^{n-1}$:\n\n$$g'(x) = 2x + 6$$"},{"type":"text","order":2,"title":"Find the stationary point","content":"To find the $x$-coordinate of the stationary (minimum) point, set the derivative equal to zero:\n\n$$2x + 6 = 0$$\n\nSolving for $x$:\n\n$$\\begin{aligned} 2x &= -6 \\\\ x &= -3 \\end{aligned}$$"},{"type":"text","order":3,"title":"Calculate the minimum value","content":"Now substitute $x = -3$ back into the original function $g(x)$ to find the minimum value of the range:\n\n$$\\begin{aligned} g(-3) &= (-3)^2 + 6(-3) - 2 \\\\ &= 9 - 18 - 2 \\\\ &= -11 \\end{aligned}$$\n\nThis confirms that the minimum point of the curve is located at $(-3, -11)$.."},{"type":"text","order":4,"title":"State the final range","content":"Since the parabola opens upwards, the minimum value of $g(x)$ is $-11$, and the function can take any value greater than or equal to this.\n\nTherefore, the range of $g(x)$ is:\n\n$$g(x) \\geq -11$$"},{"type":"text","order":5,"title":"Explain why no inverse exists","content":"For a function to have an inverse, it must be a **one-to-one** function (meaning every output $y$ corresponds to exactly one input $x$).\n\nSince $g(x) = x^2 + 6x - 2$ is a quadratic function, its graph is a parabola. For any value in its range (except the minimum value of $-11$), there are two different input values of $x$ that map to the same output $g(x)$.\n\nFor example, both $x = 0$ and $x = -6$ map to $-2$:\n$$g(0) = (0)^2 + 6(0) - 2 = -2$$\n$$g(-6) = (-6)^2 + 6(-6) - 2 = -2$$\n\nSince it is a **many-to-one** function rather than a one-to-one function, $g(x)$ does not have an inverse.\n\n### Markscheme Breakdown\n* **B1**: Explaining that $g(x)$ is a many-to-one function (or that two distinct $x$ values yield the same output value)."}]}],"generatedAt":"2026-06-09T05:33:58.514Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"93ad31de-1888-4dd3-b0a9-c3be4a8a854f","label":"b","content":"Explain why the function $g(x)$ does not have an inverse","markscheme":"- It is a many-to-one function (or every value in the range, except the minimum, is mapped from two values of $x$) B1","marks":1,"order":1,"markingRubric":{"totalMarks":1,"points":[{"id":"p1","marks":1,"description":"State that the function is many-to-one or not one-to-one.","code":"B"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:04:50.857Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"conceptual_definition","label":"Function Classification Approach","steps":[{"type":"text","order":0,"title":"Recall inverse conditions","content":"For a function to have a well-defined inverse, it must be a **one-to-one** function. This means that every unique output value in its range corresponds to exactly one unique input value in its domain."},{"type":"text","order":1,"title":"Classify the function type","content":"The given function is a quadratic function:\n\n$$g(x) = x^2 + 6x - 2, \\quad x \\in \\mathbb{R}$$\n\nBecause the domain is the set of all real numbers ($x \\in \\mathbb{R}$), the graph is a full parabola. Any quadratic function defined over its entire real domain is a **many-to-one** function, as multiple different input values of $x$ map to the same output value."},{"type":"text","order":2,"title":"Demonstrate many-to-one nature","content":"We can show that $g(x)$ is many-to-one by finding two different inputs that produce the same output. Let's choose $x = -2$ and $x = -4$. Evaluating the function at these points, we find $g(-2) = (-2)^2 + 6(-2) - 2 = -10$ and $g(-4) = (-4)^2 + 6(-4) - 2 = -10$. Since $g(-2) = g(-4) = -10$, more than one $x$-value maps to the same $y$-value."},{"type":"text","order":3,"title":"Conclude the explanation","content":"Since $g(x)$ is a **many-to-one** function over the domain $x \\in \\mathbb{R}$, it **does not have an inverse**."}]},{"id":"graphical_test","label":"Horizontal Line Test","steps":[{"type":"text","order":0,"title":"Recall inverse conditions","content":"For a function to have an inverse, it must be a **one-to-one** function. This means that each output value in the range must correspond to exactly one input value in the domain. If a function is **many-to-one**, it does not have an inverse over that domain."},{"type":"text","order":1,"title":"Visualize the graph","content":"Let's analyze the shape of the function:\n\n$$g(x) = x^2 + 6x - 2, \\quad x \\in \\mathbb{R}$$\n\nThis is a quadratic function with a positive leading coefficient, representing a parabola that opens upwards. By completing the square, we can find its vertex:\n\n$$g(x) = (x + 3)^2 - 11$$\n\nThis tells us the minimum point (vertex) of the parabola is at $(-3, -11)$."},{"type":"text","order":2,"title":"Apply horizontal line test","content":"To test if a function is one-to-one, we can apply the **Horizontal Line Test**. If any horizontal line $y = c$ (where $c > -11$) is drawn on the graph, it will intersect the parabola at two distinct points. For example, setting $g(x) = 5$ gives:\n\n$$x^2 + 6x - 2 = 5 \\implies x^2 + 6x - 7 = 0 \\implies (x + 7)(x - 1) = 0$$\n\nThis gives $x = -7$ and $x = 1$. Both of these inputs yield the same output of $5$."},{"type":"text","order":3,"title":"Conclude the proof","content":"Because a horizontal line intersects the curve more than once, $g(x)$ is a **many-to-one** function. Since it is many-to-one, it does not have an inverse over the domain $x \\in \\mathbb{R}$."}]},{"id":"numerical_counterexample","label":"Numerical Counterexample","steps":[{"type":"text","order":0,"title":"Recall inverse conditions","content":"For a function to have an inverse, it must be **one-to-one**. This means that every output value in its range must correspond to exactly one input value in its domain. If we can find two distinct inputs $x_1$ and $x_2$ such that $g(x_1) = g(x_2)$, then the function is **many-to-one** and does not have an inverse."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$g(\\htmlData{anchor=x1}{\\color{@sky}{0}}) = (\\color{@sky}{0})^2 + 6(\\color{@sky}{0}) - 2 = \\htmlData{anchor=out1}{\\color{@teal}{-2}}$"},{"latex":"$$g(\\htmlData{anchor=x2}{\\color{@amber}{-6}}) = (\\color{@amber}{-6})^2 + 6(\\color{@amber}{-6}) - 2 = \\htmlData{anchor=out2}{\\color{@teal}{-2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x1"},{"color":"amber","style":"text-color","anchor":"x2"},{"color":"teal","style":"box","anchor":"out1"},{"color":"teal","style":"box","anchor":"out2"}],"connections":[{"to":"out2","from":"out1","color":"teal","label":"same output","style":"dashed"}]},"order":1,"title":"Find a counterexample","description":"Let's find a simple output value to test, for example, when the output is $-2$. We can solve $g(x) = -2$, which simplifies to $x^2 + 6x = 0$, giving $x = 0$ and $x = -6$."},{"type":"text","order":2,"title":"Interpret the result","content":"Since two distinct inputs, $x = 0$ and $x = -6$, both map to the same output $g(0) = g(-6) = -2$, the function is **many-to-one**. Therefore, $g(x)$ does not have an inverse function over the domain of all real numbers ($x \\in \\mathbb{R}$)."}]}],"generatedAt":"2026-06-09T05:36:00.208Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"11a3f5e5-6f30-4b6a-989a-cc97f78da5c8:completing_the_square","label":"(a) Completing the Square","steps":[{"type":"text","order":0,"title":"Understand the task","content":"To find the range of the function $g(x) = x^2 + 6x - 2$, we want to find all the possible output values of $g(x)$ for any real input $x$.\n\nSince this is a quadratic function with a positive $x^2$ coefficient, its graph is a u-shaped parabola. Its range will be all values greater than or equal to its minimum value.\n\nWe can find this minimum value by rewriting the quadratic in **vertex form**, $g(x) = (x-h)^2 + k$, using the **completing the square** method."},{"type":"text","order":1,"title":"Complete the square","content":"We focus on the first two terms, $x^2 + 6x$. We halve the coefficient of $x$ (which is $6$) to get $3$, and then subtract its square ($3^2 = 9$):\n\n$$\\begin{aligned} g(x) &= x^2 + 6x - 2 \\\\ &= (x + 3)^2 - 9 - 2 \\\\ &= (x + 3)^2 - 11 \\end{aligned}$$"},{"type":"text","order":2,"title":"Find the minimum value","content":"Now that we have $g(x) = (x+3)^2 - 11$, let's analyze the expression to find its minimum value:\n\n* The squared term, $(x+3)^2$, is always non-negative because any real number squared is greater than or equal to zero:\n $$(x+3)^2 \\ge 0$$\n* The minimum possible value of this squared term is $0$, which occurs when $x = -3$.\n* Substituting $(x+3)^2 = 0$ gives the minimum value of the function:\n $$g(-3) = 0 - 11 = -11$$\n\nThus, the minimum point (vertex) of the curve is $(-3, -11)$.."},{"type":"text","order":3,"title":"State the range","content":"Since the minimum value of $g(x)$ is $-11$, the function can take any value greater than or equal to $-11$.\n\nTherefore, the range of $g(x)$ is:\n\n$$g(x) \\ge -11$$\n\n### Markscheme Breakdown\n* **M1**: Correctly completing the square to get $g(x) = (x+3)^2 - 11$\n* **M1**: Identifying the minimum point as $(-3, -11)$\n* **A1**: Writing the range as $g(x) \\ge -11$ (or $y \\ge -11$)"},{"type":"text","order":4,"title":"Explain why no inverse exists","content":"For a function to have an inverse, it must be a **one-to-one** function (meaning every output $y$ corresponds to exactly one input $x$).\n\nSince $g(x) = x^2 + 6x - 2$ is a quadratic function, its graph is a parabola. For any value in its range (except the minimum value of $-11$), there are two different input values of $x$ that map to the same output $g(x)$.\n\nFor example, both $x = 0$ and $x = -6$ map to $-2$:\n$$g(0) = (0)^2 + 6(0) - 2 = -2$$\n$$g(-6) = (-6)^2 + 6(-6) - 2 = -2$$\n\nSince it is a **many-to-one** function rather than a one-to-one function, $g(x)$ does not have an inverse.\n\n### Markscheme Breakdown\n* **B1**: Explaining that $g(x)$ is a many-to-one function (or that two distinct $x$ values yield the same output value)."}]},{"id":"11a3f5e5-6f30-4b6a-989a-cc97f78da5c8:vertex_formula","label":"(a) Using the Vertex Formula","steps":[{"type":"text","order":0,"title":"Identify the coefficients","content":"To find the vertex of the quadratic function $g(x) = x^2 + 6x - 2$, we first identify the coefficients from the general quadratic form $g(x) = ax^2 + bx + c$:\n\n* $$a = 1$$\n* $$b = 6$$\n* $$c = -2$$"},{"type":"text","order":1,"title":"Calculate vertex x-coordinate","content":"We use the vertex formula $x = -\\frac{b}{2a}$ to find the $x$-coordinate of the vertex of the parabola.\n\nSubstituting $a = 1$ and $b = 6$:\n\n$$x = -\\frac{6}{2(1)} = -3$$"},{"type":"text","order":2,"title":"Find the minimum value","content":"Now we evaluate $g(x)$ at $x = -3$ to find the corresponding $y$-coordinate of the vertex:\n\n$$\\begin{aligned} g(-3) &= (-3)^2 + 6(-3) - 2 \\\\ &= 9 - 18 - 2 \\\\ &= -11 \\end{aligned}$$\n\nSo, the vertex of the parabola is at $(-3, -11)$.."},{"type":"text","order":3,"title":"Determine the range","content":"Since the coefficient of $x^2$ is positive ($a = 1 > 0$), the parabola opens upwards.\n\nThis means the vertex represents the absolute minimum point of the function.\n\nTherefore, the range of $g(x)$ is all values greater than or equal to this minimum:\n\n$$g(x) \\geq -11$$","calculator":[{"gdc":{"expression":"x^2 + 6*x - 2","instructions":"1. Go to the Graph application.\n2. Enter the function \$f(x) = x^2 + 6x - 2\$.\n3. Use the 'Analyze Graph' or 'G-Solv' tool and select 'Minimum'.\n4. The calculator will display the vertex coordinates at \$(-3, -11)\$."},"ti84":{"expression":"X^2 + 6*X - 2","instructions":"1. Press [Y=] and enter \$Y_1 = X^2 + 6X - 2\$.\n2. Press [GRAPH] to view the curve.\n3. Press [2nd] [TRACE] (CALC) and select '3: minimum'.\n4. Set a 'Left Bound' to the left of the vertex, a 'Right Bound' to the right, and press [ENTER] on 'Guess'.\n5. The screen will display the coordinates of the minimum point at approximately \$x = -3, y = -11\$."},"desmos":{"expression":"y = x^2 + 6*x - 2","instructions":"1. Type `y = x^2 + 6x - 2` into the input bar.\n2. Click on the curve to reveal key points.\n3. Click on the gray dot at the very bottom of the parabola to see the coordinates of the minimum vertex point: \$(-3, -11)\$."},"description":"You can verify this range visually by graphing the function and finding its minimum value."}]},{"type":"text","order":4,"title":"Explain why no inverse exists","content":"For a function to have an inverse, it must be a **one-to-one** function (meaning every output $y$ corresponds to exactly one input $x$).\n\nSince $g(x) = x^2 + 6x - 2$ is a quadratic function, its graph is a parabola. For any value in its range (except the minimum value of $-11$), there are two different input values of $x$ that map to the same output $g(x)$.\n\nFor example, both $x = 0$ and $x = -6$ map to $-2$:\n$$g(0) = (0)^2 + 6(0) - 2 = -2$$\n$$g(-6) = (-6)^2 + 6(-6) - 2 = -2$$\n\nSince it is a **many-to-one** function rather than a one-to-one function, $g(x)$ does not have an inverse.\n\n### Markscheme Breakdown\n* **B1**: Explaining that $g(x)$ is a many-to-one function (or that two distinct $x$ values yield the same output value)."}]},{"id":"11a3f5e5-6f30-4b6a-989a-cc97f78da5c8:calculus","label":"(a) Using Differentiation","steps":[{"type":"text","order":0,"title":"Understand the goal","content":"We want to find the range of the quadratic function:\n\n$$g(x) = x^2 + 6x - 2$$\n\nBecause the coefficient of $x^2$ is positive ($1 > 0$), the graph of this function is a $U$-shaped parabola that opens upwards. This means the function has a global minimum point. The range will include all values greater than or equal to this minimum value:\n\n$$\\text{Range: } g(x) \\geq g(x_{\\text{min}})$$\n\nWe will use differentiation to locate this minimum point."},{"type":"text","order":1,"title":"Find the derivative","content":"Differentiate $g(x)$ term-by-term with respect to $x$ using the power rule $\\frac{d}{dx}(x^n) = n x^{n-1}$:\n\n$$g'(x) = 2x + 6$$"},{"type":"text","order":2,"title":"Find the stationary point","content":"To find the $x$-coordinate of the stationary (minimum) point, set the derivative equal to zero:\n\n$$2x + 6 = 0$$\n\nSolving for $x$:\n\n$$\\begin{aligned} 2x &= -6 \\\\ x &= -3 \\end{aligned}$$"},{"type":"text","order":3,"title":"Calculate the minimum value","content":"Now substitute $x = -3$ back into the original function $g(x)$ to find the minimum value of the range:\n\n$$\\begin{aligned} g(-3) &= (-3)^2 + 6(-3) - 2 \\\\ &= 9 - 18 - 2 \\\\ &= -11 \\end{aligned}$$\n\nThis confirms that the minimum point of the curve is located at $(-3, -11)$.."},{"type":"text","order":4,"title":"State the final range","content":"Since the parabola opens upwards, the minimum value of $g(x)$ is $-11$, and the function can take any value greater than or equal to this.\n\nTherefore, the range of $g(x)$ is:\n\n$$g(x) \\geq -11$$"},{"type":"text","order":5,"title":"Explain why no inverse exists","content":"For a function to have an inverse, it must be a **one-to-one** function (meaning every output $y$ corresponds to exactly one input $x$).\n\nSince $g(x) = x^2 + 6x - 2$ is a quadratic function, its graph is a parabola. For any value in its range (except the minimum value of $-11$), there are two different input values of $x$ that map to the same output $g(x)$.\n\nFor example, both $x = 0$ and $x = -6$ map to $-2$:\n$$g(0) = (0)^2 + 6(0) - 2 = -2$$\n$$g(-6) = (-6)^2 + 6(-6) - 2 = -2$$\n\nSince it is a **many-to-one** function rather than a one-to-one function, $g(x)$ does not have an inverse.\n\n### Markscheme Breakdown\n* **B1**: Explaining that $g(x)$ is a many-to-one function (or that two distinct $x$ values yield the same output value)."}]},{"id":"93ad31de-1888-4dd3-b0a9-c3be4a8a854f:conceptual_definition","label":"(b) Function Classification Approach","steps":[{"type":"text","order":0,"title":"Recall inverse conditions","content":"For a function to have a well-defined inverse, it must be a **one-to-one** function. This means that every unique output value in its range corresponds to exactly one unique input value in its domain."},{"type":"text","order":1,"title":"Classify the function type","content":"The given function is a quadratic function:\n\n$$g(x) = x^2 + 6x - 2, \\quad x \\in \\mathbb{R}$$\n\nBecause the domain is the set of all real numbers ($x \\in \\mathbb{R}$), the graph is a full parabola. Any quadratic function defined over its entire real domain is a **many-to-one** function, as multiple different input values of $x$ map to the same output value."},{"type":"text","order":2,"title":"Demonstrate many-to-one nature","content":"We can show that $g(x)$ is many-to-one by finding two different inputs that produce the same output. Let's choose $x = -2$ and $x = -4$. Evaluating the function at these points, we find $g(-2) = (-2)^2 + 6(-2) - 2 = -10$ and $g(-4) = (-4)^2 + 6(-4) - 2 = -10$. Since $g(-2) = g(-4) = -10$, more than one $x$-value maps to the same $y$-value."},{"type":"text","order":3,"title":"Conclude the explanation","content":"Since $g(x)$ is a **many-to-one** function over the domain $x \\in \\mathbb{R}$, it **does not have an inverse**."}]},{"id":"93ad31de-1888-4dd3-b0a9-c3be4a8a854f:graphical_test","label":"(b) Horizontal Line Test","steps":[{"type":"text","order":0,"title":"Recall inverse conditions","content":"For a function to have an inverse, it must be a **one-to-one** function. This means that each output value in the range must correspond to exactly one input value in the domain. If a function is **many-to-one**, it does not have an inverse over that domain."},{"type":"text","order":1,"title":"Visualize the graph","content":"Let's analyze the shape of the function:\n\n$$g(x) = x^2 + 6x - 2, \\quad x \\in \\mathbb{R}$$\n\nThis is a quadratic function with a positive leading coefficient, representing a parabola that opens upwards. By completing the square, we can find its vertex:\n\n$$g(x) = (x + 3)^2 - 11$$\n\nThis tells us the minimum point (vertex) of the parabola is at $(-3, -11)$."},{"type":"text","order":2,"title":"Apply horizontal line test","content":"To test if a function is one-to-one, we can apply the **Horizontal Line Test**. If any horizontal line $y = c$ (where $c > -11$) is drawn on the graph, it will intersect the parabola at two distinct points. For example, setting $g(x) = 5$ gives:\n\n$$x^2 + 6x - 2 = 5 \\implies x^2 + 6x - 7 = 0 \\implies (x + 7)(x - 1) = 0$$\n\nThis gives $x = -7$ and $x = 1$. Both of these inputs yield the same output of $5$."},{"type":"text","order":3,"title":"Conclude the proof","content":"Because a horizontal line intersects the curve more than once, $g(x)$ is a **many-to-one** function. Since it is many-to-one, it does not have an inverse over the domain $x \\in \\mathbb{R}$."}]},{"id":"93ad31de-1888-4dd3-b0a9-c3be4a8a854f:numerical_counterexample","label":"(b) Numerical Counterexample","steps":[{"type":"text","order":0,"title":"Recall inverse conditions","content":"For a function to have an inverse, it must be **one-to-one**. This means that every output value in its range must correspond to exactly one input value in its domain. If we can find two distinct inputs $x_1$ and $x_2$ such that $g(x_1) = g(x_2)$, then the function is **many-to-one** and does not have an inverse."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$g(\\htmlData{anchor=x1}{\\color{@sky}{0}}) = (\\color{@sky}{0})^2 + 6(\\color{@sky}{0}) - 2 = \\htmlData{anchor=out1}{\\color{@teal}{-2}}$"},{"latex":"$$g(\\htmlData{anchor=x2}{\\color{@amber}{-6}}) = (\\color{@amber}{-6})^2 + 6(\\color{@amber}{-6}) - 2 = \\htmlData{anchor=out2}{\\color{@teal}{-2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x1"},{"color":"amber","style":"text-color","anchor":"x2"},{"color":"teal","style":"box","anchor":"out1"},{"color":"teal","style":"box","anchor":"out2"}],"connections":[{"to":"out2","from":"out1","color":"teal","label":"same output","style":"dashed"}]},"order":1,"title":"Find a counterexample","description":"Let's find a simple output value to test, for example, when the output is $-2$. We can solve $g(x) = -2$, which simplifies to $x^2 + 6x = 0$, giving $x = 0$ and $x = -6$."},{"type":"text","order":2,"title":"Interpret the result","content":"Since two distinct inputs, $x = 0$ and $x = -6$, both map to the same output $g(0) = g(-6) = -2$, the function is **many-to-one**. Therefore, $g(x)$ does not have an inverse function over the domain of all real numbers ($x \\in \\mathbb{R}$)."}]}],"generatedAt":"2026-06-09T05:36:11.958Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":62},"walkthroughVideo":null}],"subjectCode":"a_level_maths","summaries":[{"id":"89fb776e-6c83-4fd6-9eb4-075e1c9fb82d","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"5d3e02dc-cec3-4cac-9d59-24032bd4142d","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"46c2b354-ef56-4172-8c02-d507966463e7","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"08c6df0b-f9cf-49cd-a5f5-554140455f97","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"600b05a2-86cb-49b5-9c1d-a14aabe63e11","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true}],"title":"2.2 Functions and Mappings","topicId":"0ee413bd-2885-447b-b456-d3bbd461cebc","topicSidebar":"$L3b","topicSlug":"a-level-pure-mathematics-2-2-functions-and-mappings","topicTitle":"2.2 Functions and Mappings"}]}],"$L3c","$L3d"]}]