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+ 7$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Rearranges the $x$ equation to make $t$ the subject.","code":"M"},{"id":"p2","marks":1,"description":"Substitutes their expression for $t$ into the $y$ equation.","code":"M"},{"id":"p3","marks":1,"description":"Correct cartesian equation $y = (x - 2)^2 + 3$.","code":"A"}],"finalAnswers":["y = (x - 2)^2 + 3","y = x^2 - 4x + 7"],"generatedAt":"2026-05-06T03:39:34.672Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution_from_x","label":"Substitution from x-equation","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=x-eq}{x} = \\htmlData{anchor=t-term}{t} + 2$"},{"latex":"$$\\htmlData{anchor=t-eq}{\\color{@sky}{t}} = \\htmlData{anchor=x-minus-2}{\\color{@sky}{x - 2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-eq"},{"color":"sky","style":"underline","anchor":"x-minus-2"}],"connections":[{"to":"t-eq","from":"t-term","color":"sky","label":"rearrange for t","style":"solid"}]},"order":0,"title":"Rearrange the x-equation","description":"To eliminate the parameter $t$, we start by rearranging the linear parametric equation for $x$ to make $t$ the subject. Subtract $2$ from both sides of the equation."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = (\\htmlData{anchor=t-sub}{\\color{@sky}{t}})^2 + 3$"},{"latex":"$$y = (\\htmlData{anchor=substituted}{\\color{@sky}{x - 2}})^2 + 3$"}],"highlights":[{"color":"sky","style":"box","anchor":"substituted"}],"connections":[{"to":"substituted","from":"t-sub","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into y-equation","description":"Next, substitute this expression for $t$ into the parametric equation for $y$ to get an equation in terms of $x$ and $y$ only."},{"type":"text","order":2,"title":"Simplify and state final answer","content":"You can leave your answer in this completed square form, or expand it out into a standard quadratic trinomial.\n\nExpanding the brackets:\n$$\\begin{aligned} y &= (x - 2)^2 + 3 \\\\ y &= (x^2 - 4x + 4) + 3 \\\\ y &= x^2 - 4x + 7 \\end{aligned}$$\n\nBoth $y = (x - 2)^2 + 3$ and $y = x^2 - 4x + 7$ are fully correct and will receive full marks.","calculator":[{"gdc":{"instructions":"1. Open the graphing app on your GDC.\n2. Add a parametric relation and enter x(t) = t + 2 and y(t) = t^2 + 3.\n3. Add a standard function graph and enter f(x) = (x - 2)^2 + 3.\n4. Observe that the curves overlap perfectly."},"ti84":{"instructions":"1. Change mode to parametric: Press [MODE], arrow down to 'FUNCTION' and arrow right to select 'PARAMETRIC', then press [ENTER].\n2. Press [Y=] and enter X1T = T + 2 and Y1T = T^2 + 3.\n3. Change mode back to 'FUNCTION' and press [Y=], then enter Y1 = (X - 2)^2 + 3.\n4. Press [GRAPH] and check that both graphs are identical."},"desmos":{"expression":"y = (x - 2)^2 + 3","instructions":"1. Type the parametric equation as an ordered pair in the first row: (t + 2, t^2 + 3).\n2. Set the domain of t (e.g., -10 to 10).\n3. In the second row, type the cartesian equation: y = (x - 2)^2 + 3.\n4. Observe that the second curve lies exactly on top of the first curve."},"description":"Verify your cartesian equation graphically by plotting both the parametric equations and the cartesian equation to check if they overlap."}]}]},{"id":"substitution_from_y","label":"Substitution from y-equation","steps":[{"type":"text","order":0,"title":"Identify the goal","content":"We need to find a Cartesian equation for the curve. A Cartesian equation relates $x$ and $y$ directly, without the parameter $t$. \n\nWe are given the parametric equations:\n$$x = t + 2$$\n$$y = t^2 + 3$$\n\nTo find the Cartesian equation, we will rearrange the $y$-equation to express $t$ in terms of $y$, substitute that expression into our $x$-equation, and then simplify.","highlights":[{"text":"x = t + 2, \\quad y = t^2 + 3","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=y-eq}{y} = \\htmlData{anchor=t-term}{t^2} + 3$"},{"latex":"$$\\htmlData{anchor=t2-eq}{t^2} = \\htmlData{anchor=y-minus}{y - 3}$"},{"latex":"$$\\htmlData{anchor=t-sol}{t} = \\pm\\sqrt{\\htmlData{anchor=sqrt-term}{y - 3}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-eq"},{"color":"amber","style":"text-color","anchor":"t-term"},{"color":"amber","style":"text-color","anchor":"t-sol"}],"connections":[{"to":"t2-eq","from":"y-eq","color":"sky","label":"rearrange","style":"solid"},{"to":"t-sol","from":"t-term","color":"amber","label":"square root","style":"solid"}]},"order":1,"title":"Express t using y","description":"First, we rearrange the equation for $y$ to solve for $t^2$, then take the square root of both sides."},{"type":"text","order":2,"title":"Substitute into x-equation","content":"Next, we substitute our expression for $t = \\pm\\sqrt{y - 3}$ into the equation for $x$:\n\n$$x = t + 2$$\n\n$$x = \\pm\\sqrt{y - 3} + 2$$"},{"type":"text","order":3,"title":"Isolate the radical","content":"Before squaring, we should isolate the radical term on one side of the equation. Subtract $2$ from both sides:\n\n$$x - 2 = \\pm\\sqrt{y - 3}$$"},{"type":"text","order":4,"title":"Square and simplify","content":"Now, square both sides of the equation to eliminate the radical:\n\n$$(x - 2)^2 = \\left(\\pm\\sqrt{y - 3}\\right)^2$$\n\n$$(x - 2)^2 = y - 3$$\n\nRearranging to make $y$ the subject gives the final Cartesian equation:\n\n$$y = (x - 2)^2 + 3$$ \n\nWe can also expand this to polynomial form:\n\n$$y = x^2 - 4x + 4 + 3$$\n$$y = x^2 - 4x + 7$$"}]}],"generatedAt":"2026-06-09T10:54:31.714Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"a80f3d88-bd42-4348-8241-ca20792c682f","label":"b","content":"Sketch the curve","markscheme":"- Correct shape of a parabola (u-shape) B1\n- Minimum point correctly labeled at $(2, 3)$ B1\n- $y$-intercept correctly labeled at $7$ B1\n\n![Sketch](img:21ac95be2778732e577d82233d5ee6ab_img.jpg)","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Correct quadratic curve shape (parabola).","code":"B"},{"id":"p2","marks":1,"description":"Vertex/Minimum point correctly identified at $(2, 3)$.","code":"B"},{"id":"p3","marks":1,"description":"$$y$-intercept correctly identified at $(0, 7)$.","code":"B"}],"finalAnswers":[],"generatedAt":"2026-05-06T03:39:34.672Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"cartesian","label":"Converting to Cartesian Form","steps":[{"type":"text","order":0,"title":"Eliminate the parameter","content":"To sketch the parametric curve, we can first convert the parametric equations into a single Cartesian equation by eliminating the parameter $t$.\n\nWe start with the equation for $x$:\n$$x = t + 2$$\n\nRearranging this to make $t$ the subject gives:\n$$t = x - 2$$","highlights":[{"text":"x = t + 2","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\text{Substitute } \frac{t = \text{\\htmlData{anchor=t-val}{x - 2}}}{\\phantom{\\text{sub}}} \text{ into the equation for } y = \\text{\\htmlData{anchor=y-eq}{t}}^2 + 3$"},{"latex":"$$y = (\\color{@sky}{x - 2})^2 + 3$"}],"highlights":[{"color":"sky","style":"underline","anchor":"t-val"}],"connections":[{"to":"y-eq","from":"t-val","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute to find y","description":"Substitute the expression for $t$ into the equation for $y$ to get the Cartesian equation of the curve."},{"type":"text","order":2,"title":"Identify the vertex","content":"The Cartesian equation is in vertex form:\n\n$$y = (x - 2)^2 + 3$$\n\nSince $(x - 2)^2 \\ge 0$ for all real values of $x$, the minimum value of $y$ occurs when $x - 2 = 0$, which gives $x = 2$. \n\nSubstituting $x = 2$ gives $y = 3$. Therefore, the curve is a **U-shaped parabola** with a minimum point (vertex) at:\n\n$$\\text{Minimum point} = (2, 3)$$"},{"type":"text","order":3,"title":"Find the y-intercept","content":"To find where the curve crosses the $y$-axis, set $x = 0$ in our Cartesian equation:\n\n$$\\begin{aligned} y &= (0 - 2)^2 + 3 \\\\ y &= (-2)^2 + 3 \\\\ y &= 4 + 3 \\\\ y &= 7 \\end{aligned}$$\n\nSo, the curve has a $y$-intercept at $(0, 7)$."},{"type":"text","order":4,"title":"Sketch the curve","content":"Now draw the sketch using these key features:\n\n1. Draw a smooth, symmetric U-shaped parabola.\n2. Clearly label the minimum point at $(2, 3)$.\n3. Mark the $y$-intercept at $7$ (or $(0, 7)$).\n\nHere is how the marks are awarded:\n* **B1** for the correct U-shape.\n* **B1** for correctly identifying the minimum point at $(2, 3)$.\n* **B1** for the $y$-intercept at $7$."}]},{"id":"tableofvalues","label":"Using a Table of Values","steps":[{"type":"text","order":0,"title":"Set up a table","content":"To sketch the parametric curve using a table of values, we select a few convenient values for the parameter $t$ and calculate the corresponding values for $x = t + 2$ and $y = t^2 + 3$.\n\nLet's choose $t$ values from $-3$ to $2$ to capture the key features of the curve:\n\n$$\\begin{array}{|c|c|c|c|} \n\\hline \nt & x = t + 2 & y = t^2 + 3 & (x, y) \\\\\n\\hline\n-3 & -1 & 12 & (-1, 12) \\\\\n-2 & 0 & 7 & (0, 7) \\\\\n-1 & 1 & 4 & (1, 4) \\\\\n0 & 2 & 3 & (2, 3) \\\\\n1 & 3 & 4 & (3, 4) \\\\\n2 & 4 & 7 & (4, 7) \\\\\n\\hline \n\\end{array}$$","calculator":[{"gdc":{"instructions":"1. In the Graph menu, change the type of graph to Parametric.\n2. Enter the equations:\n xt1 = t + 2\n yt1 = t^2 + 3\n3. Go to the Table menu to generate and view the calculated coordinate table."},"ti84":{"instructions":"1. Press [Mode] and change the setting on the 4th line from FUNCTION to PARAMETRIC (PAR).\n2. Press [Y=] and enter:\n X1T = T + 2\n Y1T = T^2 + 3\n3. Press [2nd] [Window] (TblSet) and set TblStart = -3 and ΔTbl = 1.\n4. Press [2nd] [Graph] (Table) to view the generated table of (X, Y) pairs."},"desmos":{"expression":"(t + 2, t^2 + 3)","instructions":"1. Click the '+' button and select 'Table'.\n2. In the first column, type 't' (or 'x1' and define custom expressions).\n3. Or, simply plot the parametric curve directly by typing: (t + 2, t^2 + 3) and setting the domain for t."},"description":"Generate the table of values using your graphing calculator's table function."}]},{"type":"text","order":1,"title":"Identify key coordinates","content":"From our table of coordinates, we can identify key points needed for the sketch:\n\n* **$y$-intercept**: This occurs when $x = 0$. Looking at our table, this corresponds to the point $\\mathbf{(0, 7)}$ (when $t = -2$).\n* **Minimum point**: We observe that the $y$-values decrease to $3$ at $t = 0$ and then begin to rise again symmetrically. Thus, the minimum point is located at $\\mathbf{(2, 3)}$."},{"type":"text","order":2,"title":"Sketch the curve","content":"Now we sketch the curve using the points we've calculated:\n\n1. Plot the points $(-1, 12)$, $(0, 7)$, $(1, 4)$, $(2, 3)$, $(3, 4)$, and $(4, 7)$ on a set of axes.\n2. Connect these points with a smooth, continuous, $U$-shaped curve (a parabola).\n3. Ensure you clearly label the coordinates of the **minimum point** at $(2, 3)$ and the **$y$-intercept** at $(0, 7)$ (or label $7$ on the $y$-axis).\n\n*Note on marks: You earn 1 mark for the correct $U$-shaped parabola, 1 mark for the correct minimum point, and 1 mark for the correct $y$-intercept.*"}]}],"generatedAt":"2026-06-09T10:55:04.826Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"d64ae52d-2479-4d7e-87e2-4e10b7070509:substitution_from_x","label":"(a) Substitution from x-equation","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=x-eq}{x} = \\htmlData{anchor=t-term}{t} + 2$"},{"latex":"$$\\htmlData{anchor=t-eq}{\\color{@sky}{t}} = \\htmlData{anchor=x-minus-2}{\\color{@sky}{x - 2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-eq"},{"color":"sky","style":"underline","anchor":"x-minus-2"}],"connections":[{"to":"t-eq","from":"t-term","color":"sky","label":"rearrange for t","style":"solid"}]},"order":0,"title":"Rearrange the x-equation","description":"To eliminate the parameter $t$, we start by rearranging the linear parametric equation for $x$ to make $t$ the subject. Subtract $2$ from both sides of the equation."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = (\\htmlData{anchor=t-sub}{\\color{@sky}{t}})^2 + 3$"},{"latex":"$$y = (\\htmlData{anchor=substituted}{\\color{@sky}{x - 2}})^2 + 3$"}],"highlights":[{"color":"sky","style":"box","anchor":"substituted"}],"connections":[{"to":"substituted","from":"t-sub","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into y-equation","description":"Next, substitute this expression for $t$ into the parametric equation for $y$ to get an equation in terms of $x$ and $y$ only."},{"type":"text","order":2,"title":"Simplify and state final answer","content":"You can leave your answer in this completed square form, or expand it out into a standard quadratic trinomial.\n\nExpanding the brackets:\n$$\\begin{aligned} y &= (x - 2)^2 + 3 \\\\ y &= (x^2 - 4x + 4) + 3 \\\\ y &= x^2 - 4x + 7 \\end{aligned}$$\n\nBoth $y = (x - 2)^2 + 3$ and $y = x^2 - 4x + 7$ are fully correct and will receive full marks.","calculator":[{"gdc":{"instructions":"1. Open the graphing app on your GDC.\n2. Add a parametric relation and enter x(t) = t + 2 and y(t) = t^2 + 3.\n3. Add a standard function graph and enter f(x) = (x - 2)^2 + 3.\n4. Observe that the curves overlap perfectly."},"ti84":{"instructions":"1. Change mode to parametric: Press [MODE], arrow down to 'FUNCTION' and arrow right to select 'PARAMETRIC', then press [ENTER].\n2. Press [Y=] and enter X1T = T + 2 and Y1T = T^2 + 3.\n3. Change mode back to 'FUNCTION' and press [Y=], then enter Y1 = (X - 2)^2 + 3.\n4. Press [GRAPH] and check that both graphs are identical."},"desmos":{"expression":"y = (x - 2)^2 + 3","instructions":"1. Type the parametric equation as an ordered pair in the first row: (t + 2, t^2 + 3).\n2. Set the domain of t (e.g., -10 to 10).\n3. In the second row, type the cartesian equation: y = (x - 2)^2 + 3.\n4. Observe that the second curve lies exactly on top of the first curve."},"description":"Verify your cartesian equation graphically by plotting both the parametric equations and the cartesian equation to check if they overlap."}]}]},{"id":"d64ae52d-2479-4d7e-87e2-4e10b7070509:substitution_from_y","label":"(a) Substitution from y-equation","steps":[{"type":"text","order":0,"title":"Identify the goal","content":"We need to find a Cartesian equation for the curve. A Cartesian equation relates $x$ and $y$ directly, without the parameter $t$. \n\nWe are given the parametric equations:\n$$x = t + 2$$\n$$y = t^2 + 3$$\n\nTo find the Cartesian equation, we will rearrange the $y$-equation to express $t$ in terms of $y$, substitute that expression into our $x$-equation, and then simplify.","highlights":[{"text":"x = t + 2, \\quad y = t^2 + 3","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=y-eq}{y} = \\htmlData{anchor=t-term}{t^2} + 3$"},{"latex":"$$\\htmlData{anchor=t2-eq}{t^2} = \\htmlData{anchor=y-minus}{y - 3}$"},{"latex":"$$\\htmlData{anchor=t-sol}{t} = \\pm\\sqrt{\\htmlData{anchor=sqrt-term}{y - 3}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-eq"},{"color":"amber","style":"text-color","anchor":"t-term"},{"color":"amber","style":"text-color","anchor":"t-sol"}],"connections":[{"to":"t2-eq","from":"y-eq","color":"sky","label":"rearrange","style":"solid"},{"to":"t-sol","from":"t-term","color":"amber","label":"square root","style":"solid"}]},"order":1,"title":"Express t using y","description":"First, we rearrange the equation for $y$ to solve for $t^2$, then take the square root of both sides."},{"type":"text","order":2,"title":"Substitute into x-equation","content":"Next, we substitute our expression for $t = \\pm\\sqrt{y - 3}$ into the equation for $x$:\n\n$$x = t + 2$$\n\n$$x = \\pm\\sqrt{y - 3} + 2$$"},{"type":"text","order":3,"title":"Isolate the radical","content":"Before squaring, we should isolate the radical term on one side of the equation. Subtract $2$ from both sides:\n\n$$x - 2 = \\pm\\sqrt{y - 3}$$"},{"type":"text","order":4,"title":"Square and simplify","content":"Now, square both sides of the equation to eliminate the radical:\n\n$$(x - 2)^2 = \\left(\\pm\\sqrt{y - 3}\\right)^2$$\n\n$$(x - 2)^2 = y - 3$$\n\nRearranging to make $y$ the subject gives the final Cartesian equation:\n\n$$y = (x - 2)^2 + 3$$ \n\nWe can also expand this to polynomial form:\n\n$$y = x^2 - 4x + 4 + 3$$\n$$y = x^2 - 4x + 7$$"}]},{"id":"a80f3d88-bd42-4348-8241-ca20792c682f:cartesian","label":"(b) Converting to Cartesian Form","steps":[{"type":"text","order":0,"title":"Eliminate the parameter","content":"To sketch the parametric curve, we can first convert the parametric equations into a single Cartesian equation by eliminating the parameter $t$.\n\nWe start with the equation for $x$:\n$$x = t + 2$$\n\nRearranging this to make $t$ the subject gives:\n$$t = x - 2$$","highlights":[{"text":"x = t + 2","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\text{Substitute } \frac{t = \text{\\htmlData{anchor=t-val}{x - 2}}}{\\phantom{\\text{sub}}} \text{ into the equation for } y = \\text{\\htmlData{anchor=y-eq}{t}}^2 + 3$"},{"latex":"$$y = (\\color{@sky}{x - 2})^2 + 3$"}],"highlights":[{"color":"sky","style":"underline","anchor":"t-val"}],"connections":[{"to":"y-eq","from":"t-val","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute to find y","description":"Substitute the expression for $t$ into the equation for $y$ to get the Cartesian equation of the curve."},{"type":"text","order":2,"title":"Identify the vertex","content":"The Cartesian equation is in vertex form:\n\n$$y = (x - 2)^2 + 3$$\n\nSince $(x - 2)^2 \\ge 0$ for all real values of $x$, the minimum value of $y$ occurs when $x - 2 = 0$, which gives $x = 2$. \n\nSubstituting $x = 2$ gives $y = 3$. Therefore, the curve is a **U-shaped parabola** with a minimum point (vertex) at:\n\n$$\\text{Minimum point} = (2, 3)$$"},{"type":"text","order":3,"title":"Find the y-intercept","content":"To find where the curve crosses the $y$-axis, set $x = 0$ in our Cartesian equation:\n\n$$\\begin{aligned} y &= (0 - 2)^2 + 3 \\\\ y &= (-2)^2 + 3 \\\\ y &= 4 + 3 \\\\ y &= 7 \\end{aligned}$$\n\nSo, the curve has a $y$-intercept at $(0, 7)$."},{"type":"text","order":4,"title":"Sketch the curve","content":"Now draw the sketch using these key features:\n\n1. Draw a smooth, symmetric U-shaped parabola.\n2. Clearly label the minimum point at $(2, 3)$.\n3. Mark the $y$-intercept at $7$ (or $(0, 7)$).\n\nHere is how the marks are awarded:\n* **B1** for the correct U-shape.\n* **B1** for correctly identifying the minimum point at $(2, 3)$.\n* **B1** for the $y$-intercept at $7$."}]},{"id":"a80f3d88-bd42-4348-8241-ca20792c682f:tableofvalues","label":"(b) Using a Table of Values","steps":[{"type":"text","order":0,"title":"Set up a table","content":"To sketch the parametric curve using a table of values, we select a few convenient values for the parameter $t$ and calculate the corresponding values for $x = t + 2$ and $y = t^2 + 3$.\n\nLet's choose $t$ values from $-3$ to $2$ to capture the key features of the curve:\n\n$$\\begin{array}{|c|c|c|c|} \n\\hline \nt & x = t + 2 & y = t^2 + 3 & (x, y) \\\\\n\\hline\n-3 & -1 & 12 & (-1, 12) \\\\\n-2 & 0 & 7 & (0, 7) \\\\\n-1 & 1 & 4 & (1, 4) \\\\\n0 & 2 & 3 & (2, 3) \\\\\n1 & 3 & 4 & (3, 4) \\\\\n2 & 4 & 7 & (4, 7) \\\\\n\\hline \n\\end{array}$$","calculator":[{"gdc":{"instructions":"1. In the Graph menu, change the type of graph to Parametric.\n2. Enter the equations:\n xt1 = t + 2\n yt1 = t^2 + 3\n3. Go to the Table menu to generate and view the calculated coordinate table."},"ti84":{"instructions":"1. Press [Mode] and change the setting on the 4th line from FUNCTION to PARAMETRIC (PAR).\n2. Press [Y=] and enter:\n X1T = T + 2\n Y1T = T^2 + 3\n3. Press [2nd] [Window] (TblSet) and set TblStart = -3 and ΔTbl = 1.\n4. Press [2nd] [Graph] (Table) to view the generated table of (X, Y) pairs."},"desmos":{"expression":"(t + 2, t^2 + 3)","instructions":"1. Click the '+' button and select 'Table'.\n2. In the first column, type 't' (or 'x1' and define custom expressions).\n3. Or, simply plot the parametric curve directly by typing: (t + 2, t^2 + 3) and setting the domain for t."},"description":"Generate the table of values using your graphing calculator's table function."}]},{"type":"text","order":1,"title":"Identify key coordinates","content":"From our table of coordinates, we can identify key points needed for the sketch:\n\n* **$y$-intercept**: This occurs when $x = 0$. Looking at our table, this corresponds to the point $\\mathbf{(0, 7)}$ (when $t = -2$).\n* **Minimum point**: We observe that the $y$-values decrease to $3$ at $t = 0$ and then begin to rise again symmetrically. Thus, the minimum point is located at $\\mathbf{(2, 3)}$."},{"type":"text","order":2,"title":"Sketch the curve","content":"Now we sketch the curve using the points we've calculated:\n\n1. Plot the points $(-1, 12)$, $(0, 7)$, $(1, 4)$, $(2, 3)$, $(3, 4)$, and $(4, 7)$ on a set of axes.\n2. Connect these points with a smooth, continuous, $U$-shaped curve (a parabola).\n3. Ensure you clearly label the coordinates of the **minimum point** at $(2, 3)$ and the **$y$-intercept** at $(0, 7)$ (or label $7$ on the $y$-axis).\n\n*Note on marks: You earn 1 mark for the correct $U$-shaped parabola, 1 mark for the correct minimum point, and 1 mark for the correct $y$-intercept.*"}]}],"generatedAt":"2026-06-09T10:55:04.876Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":57},"walkthroughVideo":null},{"id":"100c6223-1680-4a86-990e-d2ad9bf0528d","revisionId":"e0f47d68-4ddb-4bb2-843a-fe7b9eb8b33a","questionNumber":null,"questionType":"LA","specification":"A curve $K$ has parametric equations\n\n$$x = 3 \\sin t + 2, \\quad y = 2 \\cos^2 t - 5 \\quad 0 \\leq t \\leq 2\\pi$$","difficulty":1,"difficultyBand":"medium","totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"d4ca8f1f-5461-4650-9ac2-fa339b43d9ee","label":"a","content":"Find an equation for $K$ in the form $y = f(x)$ and state the domain for which the curve is defined.","markscheme":"- $\\sin t = \\frac{x-2}{3} \\Rightarrow \\sin^2 t = \\frac{(x-2)^2}{9}$ M1\n- $y = 2(1 - \\sin^2 t) - 5 = 2(1 - \\frac{(x-2)^2}{9}) - 5 \\Rightarrow y = -3 - \\frac{2}{9}(x-2)^2$ A1\n- Domain: $-1 \\leq \\sin t \\leq 1 \\Rightarrow -3 \\leq 3\\sin t \\leq 3 \\Rightarrow -1 \\leq x \\leq 5$ B1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Attempt to express $\\sin^2 t$ in terms of $x$ and substitute into the equation for $y$ using the identity $\\cos^2 t + \\sin^2 t = 1$.","code":"M"},{"id":"p2","marks":1,"description":"Correct Cartesian equation in the form $y = f(x)$.","code":"A"},{"id":"p3","marks":1,"description":"Correct domain $-1 \\leq x \\leq 5$ identified from the range of the $x$ parametric equation.","code":"B"}],"finalAnswers":["$$y = -3 - \\frac{2}{9}(x-2)^2, \\quad -1 \\leq x \\leq 5$"],"generatedAt":"2026-05-06T14:05:21.991Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Rearrange for sin(t)","content":"To express $y$ in terms of $x$, we begin by isolating the trigonometric term in our equation for $x$.\n\nThe parametric equation for $x$ is given by:\n$$x = 3 \\sin t + 2$$\n\nSubtract $2$ from both sides and divide by $3$ to find an expression for $\\sin t$:\n$$\\sin t = \\frac{x - 2}{3}$$\n\nSquaring both sides gives us our expression for $\\sin^2 t$:\n$$\\sin^2 t = \\frac{(x - 2)^2}{9}$$"},{"type":"text","order":1,"title":"Rewrite y using identity","content":"Now we look at our parametric equation for $y$:\n$$y = 2 \\cos^2 t - 5$$\n\nRecall the fundamental trigonometric identity relating sine and cosine:\n$$\\cos^2 t + \\sin^2 t = 1 \\implies \\cos^2 t = 1 - \\sin^2 t$$\n\nSubstituting this identity into our equation for $y$ yields:\n$$y = 2(1 - \\sin^2 t) - 5$$\n\nSimplify this expression:\n$$y = 2 - 2\\sin^2 t - 5$$\n$$y = -3 - 2\\sin^2 t$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sin^2 t = \\htmlData{anchor=sin-val}{\\color{@sky}{\\frac{(x - 2)^2}{9}}}$"},{"latex":"$$y = -3 - 2\\htmlData{anchor=sin-sub}{\\color{@sky}{\\sin^2 t}}$"},{"latex":"$$y = -3 - \\frac{2}{9}(x - 2)^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-val"},{"color":"sky","style":"text-color","anchor":"sin-sub"}],"connections":[{"to":"sin-sub","from":"sin-val","color":"sky","label":"substitute","style":"solid"}]},"order":2,"title":"Substitute and simplify","description":"Now, we substitute our expression for $\\sin^2 t$ in terms of $x$ into our rewritten equation for $y$."},{"type":"text","order":3,"title":"Determine the domain","content":"To complete the solution, we must define the domain for $x$. The domain of the curve is determined by the range of the parametric variable $t$, where $0 \\leq t \\leq 2\\pi$.\n\nOver this interval, the value of the sine function is bounded as follows:\n$$-1 \\leq \\sin t \\leq 1$$\n\nTo find the boundaries of $x$, we multiply the inequality by $3$:\n$$-3 \\leq 3\\sin t \\leq 3$$\n\nThen, we add $2$ to all parts of the inequality:\n$$-3 + 2 \\leq 3\\sin t + 2 \\leq 3 + 2$$\n$$-1 \\leq x \\leq 5$$\n\nTherefore, the domain for the curve is $[-1, 5]$."}]},{"id":"pythagorean_identity","label":"Pythagorean Identity Method","steps":[{"type":"text","order":0,"title":"Rearrange the x equation","content":"To eliminate the parameter $t$, we first rearrange the parametric equation for $x$ to express $\\sin t$ and then $\\sin^2 t$ in terms of $x$.\n\nStarting with:\n$$x = 3 \\sin t + 2$$\n\nSubtract $2$ from both sides:\n$$x - 2 = 3 \\sin t$$\n\nDivide by $3$:\n$$\\sin t = \\frac{x - 2}{3}$$\n\nSquaring both sides gives:\n$$\\sin^2 t = \\frac{(x - 2)^2}{9}$$"},{"type":"text","order":1,"title":"Rearrange the y equation","content":"Next, we rearrange the parametric equation for $y$ to express $\\cos^2 t$ in terms of $y$.\n\nStarting with:\n$$y = 2 \\cos^2 t - 5$$\n\nAdd $5$ to both sides:\n$$y + 5 = 2 \\cos^2 t$$\n\nDivide by $2$:\n$$\\cos^2 t = \\frac{y + 5}{2}$$\n\nThis gives us our expression for $\\cos^2 t$ directly."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=identity}{\\color{@purple}{\\sin^2 t + \\cos^2 t = 1}}$"},{"latex":"$$\\htmlData{anchor=sub}{\\color{@sky}{\\frac{(x - 2)^2}{9}}} + \\htmlData{anchor=suby}{\\color{@amber}{\\frac{y + 5}{2}}} = 1$"}],"highlights":[{"color":"purple","style":"underline","anchor":"identity"},{"color":"sky","style":"box","anchor":"sub"},{"color":"amber","style":"box","anchor":"suby"}],"connections":[{"to":"sub","from":"identity","color":"purple","label":"substitute","style":"dashed"}]},"order":2,"title":"Apply the Pythagorean identity","description":"We now use the fundamental Pythagorean identity $\\sin^2 t + \\cos^2 t = 1$ to link our two expressions and eliminate $t$."},{"type":"text","order":3,"title":"Solve for y","content":"We can now rearrange this equation to express $y$ in the form $f(x)$:\n\n$$\\frac{y + 5}{2} = 1 - \\frac{(x - 2)^2}{9}$$\n\nMultiply both sides by $2$:\n$$y + 5 = 2\\left(1 - \\frac{(x - 2)^2}{9}\\right)$$\n\n$$y + 5 = 2 - \\frac{2}{9}(x - 2)^2$$\n\nSubtract $5$ from both sides:\n$$y = -3 - \\frac{2}{9}(x - 2)^2$$"},{"type":"text","order":4,"title":"Determine the domain","content":"To find the domain of $f(x)$, we use the restriction on the sine function. Since $\\sin t$ is defined for all real $t$, its range is:\n$$-1 \\le \\sin t \\le 1$$\n\nSubstitute $\\sin t = \\frac{x - 2}{3}$ into the inequality:\n$$-1 \\le \\frac{x - 2}{3} \\le 1$$\n\nMultiply the entire inequality by $3$:\n$$-3 \\le x - 2 \\le 3$$\n\nAdd $2$ to all parts:\n$$-1 \\le x \\le 5$$\n\nThus, the domain for which the curve is defined is $[-1, 5]$."}]}],"generatedAt":"2026-06-09T03:19:56.386Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"add0cbb1-5b08-467e-aad9-ac56e8a1b982","label":"b","content":"Sketch the curve","markscheme":"- Negative (inverted) parabola shape B1\n- Vertex at $(2, -3)$ B1\n- Curve starts at $(-1, -5)$ and ends at $(5, -5)$ B1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Correct shape for a negative parabola within the defined domain.","code":"B"},{"id":"p2","marks":1,"description":"Correct coordinates for the vertex $(2, -3)$ indicated on the sketch.","code":"B"},{"id":"p3","marks":1,"description":"Correct endpoints at $(-1, -5)$ and $(5, -5)$ shown clearly.","code":"B"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:05:21.991Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"cartesian_conversion","label":"Cartesian Equation Method","steps":[{"type":"text","order":0,"title":"Express sin(t) in terms of x","content":"To find the Cartesian equation of the curve, we first isolate the trigonometric term $\\sin t$ from the parametric equation for $x$:\n\n$$x = 3 \\sin t + 2$$\n\nSubtracting $2$ and then dividing by $3$ gives:\n\n$$\\sin t = \\frac{x - 2}{3}$$"},{"type":"text","order":1,"title":"Apply trigonometric identity to y","content":"Next, we use the fundamental trigonometric identity:\n\n$$\\cos^2 t = 1 - \\sin^2 t$$\n\nSubstituting this identity into our parametric equation for $y$:\n\n$$y = 2 \\cos^2 t - 5$$\n\n$$y = 2(1 - \\sin^2 t) - 5$$\n\n$$y = -2\\sin^2 t - 3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sin t = \\htmlData{anchor=sin-expr}{\\color{@sky}{\\frac{x - 2}{3}}}$"},{"latex":"$$y = -2 (\\htmlData{anchor=sin-sub}{\\color{@sky}{\\sin t}})^2 - 3$"},{"latex":"$$y = -\\frac{2}{9}(x - 2)^2 - 3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-expr"},{"color":"sky","style":"text-color","anchor":"sin-sub"}],"connections":[{"to":"sin-sub","from":"sin-expr","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Substitute to get Cartesian equation","description":"Now, we substitute our expression for $\\sin t$ into the simplified equation for $y$ to get the final Cartesian equation."},{"type":"text","order":3,"title":"Determine domain and endpoints","content":"Since $0 \\le t \\le 2\\pi$, the range of $\\sin t$ is restricted:\n\n$$-1 \\le \\sin t \\le 1$$\n\nWe can find the domain restriction for $x$ by substituting these limits into $x = 3 \\sin t + 2$:\n\n- Minimum value: when $\\sin t = -1$, $x = 3(-1) + 2 = -1$\n- Maximum value: when $\\sin t = 1$, $x = 3(1) + 2 = 5$\n\nSo, the domain restriction is $-1 \\le x \\le 5$.\n\nAt both endpoints $x = -1$ and $x = 5$, the corresponding $y$-value is:\n\n$$y = -\\frac{2}{9}(\\pm 3)^2 - 3 = -2 - 3 = -5$$\n\nThis gives the end coordinates of our curve: $(-1, -5)$ and $(5, -5)$."},{"type":"text","order":4,"title":"Identify the vertex","content":"The Cartesian equation is in the vertex form of a quadratic, $y = a(x-h)^2 + k$:\n\n$$y = -\\frac{2}{9}(x - 2)^2 - 3$$\n\n- Since the coefficient $-\\frac{2}{9}$ is negative, the parabola is **inverted** (opening downwards).\n- The vertex $(h, k)$ is located at $(2, -3)$, which is the maximum point of this inverted parabola."},{"type":"text","order":5,"title":"Sketch the curve","content":"Now we sketch the curve using our key features:\n\n1. Plot the vertex at $(2, -3)$ as a maximum point.\n2. Plot the endpoints at $(-1, -5)$ and $(5, -5)$.\n3. Draw a smooth, symmetric, inverted parabola shape connecting these points. Ensure your curve does not extend beyond $x = -1$ or $x = 5$.","calculator":[{"gdc":{"instructions":"1. Change your graphing mode to Parametric.\n2. Enter the equations: x(t) = 3*sin(t) + 2 and y(t) = 2*cos(t)^2 - 5.\n3. Set the domain variable t from 0 to 6.28.\n4. Set an appropriate window size to view the interval [-1, 5] on the x-axis and [-6, -2] on the y-axis."},"ti84":{"instructions":"1. Press [MODE], arrow down to 'FUNC', change it to 'PARAMETRIC' (PAR), and press [ENTER].\n2. Press [Y=] and enter:\n X1T = 3 * sin(T) + 2\n Y1T = 2 * (cos(T))^2 - 5\n3. Press [WINDOW] and set:\n Tmin = 0, Tmax = 6.28, Tstep = 0.1\n Xmin = -3, Xmax = 7, Xscl = 1\n Ymin = -7, Ymax = -1, Yscl = 1\n4. Press [GRAPH] to view the curve."},"desmos":{"expression":"(3*sin(t) + 2, 2*(cos(t))^2 - 5)","instructions":"1. Type the coordinate pair representing the parametric equations in a single line:\n (3*sin(t) + 2, 2*(cos(t))^2 - 5)\n2. Set the parameter interval beneath it to: 0 <= t <= 2pi."},"description":"Visualize the parametric curve to verify its shape, vertex, and domain endpoints."}]}]},{"id":"point_plotting","label":"Point-by-Point Plotting","steps":[{"type":"text","order":0,"title":"Set up a coordinate table","content":"To sketch the parametric curve using a point-by-point plotting approach, we calculate the coordinates $(x, y)$ for key values of $t$ in the interval $0 \\le t \\le 2\\pi$.\n\nLet's choose the standard angles:\n$$t = 0, \\quad \\frac{\\pi}{2}, \\quad \\pi, \\quad \\frac{3\\pi}{2}, \\quad 2\\pi$$\n\nOur parametric equations are:\n$$x = 3 \\sin t + 2$$\n$$y = 2 \\cos^2 t - 5$$"},{"type":"text","order":1,"title":"Evaluate points for t","content":"Let's calculate the coordinates $(x, y)$ for each chosen value of $t$:\n\n1. **For $t = 0$:**\n $$x = 3 \\sin(0) + 2 = 2$$\n $$y = 2 \\cos^2(0) - 5 = 2(1)^2 - 5 = -3$$\n Point: $(2, -3)$\n\n2. **For $t = \\frac{\\pi}{2}$:**\n $$x = 3 \\sin\\left(\\frac{\\pi}{2}\\right) + 2 = 3(1) + 2 = 5$$\n $$y = 2 \\cos^2\\left(\\frac{\\pi}{2}\\right) - 5 = 2(0)^2 - 5 = -5$$\n Point: $(5, -5)$\n\n3. **For $t = \\pi$:**\n $$x = 3 \\sin(\\pi) + 2 = 2$$\n $$y = 2 \\cos^2(\\pi) - 5 = 2(-1)^2 - 5 = -3$$\n Point: $(2, -3)$\n\n4. **For $t = \\frac{3\\pi}{2}$:**\n $$x = 3 \\sin\\left(\\frac{3\\pi}{2}\\right) + 2 = 3(-1) + 2 = -1$$\n $$y = 2 \\cos^2\\left(\\frac{3\\pi}{2}\\right) - 5 = 2(0)^2 - 5 = -5$$\n Point: $(-1, -5)$\n\n5. **For $t = 2\\pi$:**\n $$x = 3 \\sin(2\\pi) + 2 = 2$$\n $$y = 2 \\cos^2(2\\pi) - 5 = 2(1)^2 - 5 = -3$$\n Point: $(2, -3)$"},{"type":"text","order":2,"title":"Summarise coordinate table","content":"Let's compile these evaluated points into a clear summary table:\n\n$$\\begin{array}{|c|c|c|c|} \\hline t & x = 3\\sin t + 2 & y = 2\\cos^2 t - 5 & (x, y) \\\\ \\hline 0 & 2 & -3 & (2, -3) \\\\ \\hline \\frac{\\pi}{2} & 5 & -5 & (5, -5) \\\\ \\hline \\pi & 2 & -3 & (2, -3) \\\\ \\hline \\frac{3\\pi}{2} & -1 & -5 & (-1, -5) \\\\ \\hline 2\\pi & 2 & -3 & (2, -3) \\\\ \\hline \\end{array}$$\n\nNotice that as $t$ goes from $0$ to $2\\pi$:\n- The curve starts at the vertex $(2, -3)$.\n- It moves to the right endpoint $(5, -5)$ at $t = \\frac{\\pi}{2}$.\n- It returns to the vertex $(2, -3)$ at $t = \\pi$.\n- It moves to the left endpoint $(-1, -5)$ at $t = \\frac{3\\pi}{2}$.\n- It returns to the vertex $(2, -3)$ at $t = 2\\pi$."},{"type":"text","order":3,"title":"Analyze the shape","content":"Let's understand the underlying algebraic shape to draw the sketch accurately.\n\nFrom our table, we can see:\n- The maximum value of $y$ is $-3$, which occurs when $x = 2$. This indicates a peak or vertex at $(2, -3)$.\n- The minimum value of $y$ is $-5$, which occurs at the boundary points $x = -1$ and $x = 5$.\n- The coordinates are symmetric about the vertical line $x = 2$.\n\nThis behavior matches an inverted (negative) parabola opening downwards, with its vertex at $(2, -3)$ and endpoints at $(-1, -5)$ and $(5, -5)$."},{"type":"text","order":4,"title":"Verify with GDC","content":"We can easily verify the plotted path and the endpoints of the curve using a graphing calculator.","calculator":[{"gdc":{"instructions":"1. Set your graphing calculator's mode to Parametric equations.\n2. Define X1(t) = 3*sin(t) + 2 and Y1(t) = 2*cos(t)^2 - 5.\n3. Adjust the interval for t to [0, 6.28].\n4. Set a suitable window view (e.g., x from -2 to 6, y from -6 to -2).\n5. Plot the graph to verify the shape."},"ti84":{"instructions":"1. Press [MODE] and change the graphing mode to PAR (parametric).\n2. Press [Y=].\n3. Set X1T = 3 * sin(T) + 2 and Y1T = 2 * (cos(T))^2 - 5.\n4. Press [WINDOW] and set Tmin = 0, Tmax = 6.28, and Tstep = 0.1.\n5. Set Xmin = -2, Xmax = 6, Ymin = -6, Ymax = -2.\n6. Press [GRAPH] to view the curve."},"desmos":{"expression":"(3*sin(t) + 2, 2*cos(t)^2 - 5)","instructions":"1. Enter the parametric coordinate pair directly into an input row: (3*sin(t) + 2, 2*cos(t)^2 - 5).\n2. Set the parameter domain bounds next to the expression as 0 <= t <= 2*pi.\n3. Adjust the graph viewport to focus on the region around x in [-2, 6] and y in [-6, -2]."},"description":"Graph the parametric equations to visualize the negative parabola shape."}]},{"type":"text","order":5,"title":"Sketch the curve","content":"Now, sketch the curve on your axes:\n\n1. Plot the vertex at $(2, -3)$ as the highest point.\n2. Plot the left endpoint at $(-1, -5)$ and the right endpoint at $(5, -5)$.\n3. Draw a smooth, symmetric, downward-opening parabolic arc connecting $(-1, -5)$ to $(2, -3)$ and then to $(5, -5)$.\n\n**Marks awarded for:**\n- **B1**: Drawing an inverted (negative) parabola shape.\n- **B1**: Showing the vertex clearly located at $(2, -3)$.\n- **B1**: Showing the curve starting/ending at $(-1, -5)$ and $(5, -5)$ with no parts of the curve drawn outside this domain."}]}],"generatedAt":"2026-06-09T03:23:08.136Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"d4ca8f1f-5461-4650-9ac2-fa339b43d9ee:substitution","label":"(a) Substitution Method","steps":[{"type":"text","order":0,"title":"Rearrange for sin(t)","content":"To express $y$ in terms of $x$, we begin by isolating the trigonometric term in our equation for $x$.\n\nThe parametric equation for $x$ is given by:\n$$x = 3 \\sin t + 2$$\n\nSubtract $2$ from both sides and divide by $3$ to find an expression for $\\sin t$:\n$$\\sin t = \\frac{x - 2}{3}$$\n\nSquaring both sides gives us our expression for $\\sin^2 t$:\n$$\\sin^2 t = \\frac{(x - 2)^2}{9}$$"},{"type":"text","order":1,"title":"Rewrite y using identity","content":"Now we look at our parametric equation for $y$:\n$$y = 2 \\cos^2 t - 5$$\n\nRecall the fundamental trigonometric identity relating sine and cosine:\n$$\\cos^2 t + \\sin^2 t = 1 \\implies \\cos^2 t = 1 - \\sin^2 t$$\n\nSubstituting this identity into our equation for $y$ yields:\n$$y = 2(1 - \\sin^2 t) - 5$$\n\nSimplify this expression:\n$$y = 2 - 2\\sin^2 t - 5$$\n$$y = -3 - 2\\sin^2 t$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sin^2 t = \\htmlData{anchor=sin-val}{\\color{@sky}{\\frac{(x - 2)^2}{9}}}$"},{"latex":"$$y = -3 - 2\\htmlData{anchor=sin-sub}{\\color{@sky}{\\sin^2 t}}$"},{"latex":"$$y = -3 - \\frac{2}{9}(x - 2)^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-val"},{"color":"sky","style":"text-color","anchor":"sin-sub"}],"connections":[{"to":"sin-sub","from":"sin-val","color":"sky","label":"substitute","style":"solid"}]},"order":2,"title":"Substitute and simplify","description":"Now, we substitute our expression for $\\sin^2 t$ in terms of $x$ into our rewritten equation for $y$."},{"type":"text","order":3,"title":"Determine the domain","content":"To complete the solution, we must define the domain for $x$. The domain of the curve is determined by the range of the parametric variable $t$, where $0 \\leq t \\leq 2\\pi$.\n\nOver this interval, the value of the sine function is bounded as follows:\n$$-1 \\leq \\sin t \\leq 1$$\n\nTo find the boundaries of $x$, we multiply the inequality by $3$:\n$$-3 \\leq 3\\sin t \\leq 3$$\n\nThen, we add $2$ to all parts of the inequality:\n$$-3 + 2 \\leq 3\\sin t + 2 \\leq 3 + 2$$\n$$-1 \\leq x \\leq 5$$\n\nTherefore, the domain for the curve is $[-1, 5]$."}]},{"id":"d4ca8f1f-5461-4650-9ac2-fa339b43d9ee:pythagorean_identity","label":"(a) Pythagorean Identity Method","steps":[{"type":"text","order":0,"title":"Rearrange the x equation","content":"To eliminate the parameter $t$, we first rearrange the parametric equation for $x$ to express $\\sin t$ and then $\\sin^2 t$ in terms of $x$.\n\nStarting with:\n$$x = 3 \\sin t + 2$$\n\nSubtract $2$ from both sides:\n$$x - 2 = 3 \\sin t$$\n\nDivide by $3$:\n$$\\sin t = \\frac{x - 2}{3}$$\n\nSquaring both sides gives:\n$$\\sin^2 t = \\frac{(x - 2)^2}{9}$$"},{"type":"text","order":1,"title":"Rearrange the y equation","content":"Next, we rearrange the parametric equation for $y$ to express $\\cos^2 t$ in terms of $y$.\n\nStarting with:\n$$y = 2 \\cos^2 t - 5$$\n\nAdd $5$ to both sides:\n$$y + 5 = 2 \\cos^2 t$$\n\nDivide by $2$:\n$$\\cos^2 t = \\frac{y + 5}{2}$$\n\nThis gives us our expression for $\\cos^2 t$ directly."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=identity}{\\color{@purple}{\\sin^2 t + \\cos^2 t = 1}}$"},{"latex":"$$\\htmlData{anchor=sub}{\\color{@sky}{\\frac{(x - 2)^2}{9}}} + \\htmlData{anchor=suby}{\\color{@amber}{\\frac{y + 5}{2}}} = 1$"}],"highlights":[{"color":"purple","style":"underline","anchor":"identity"},{"color":"sky","style":"box","anchor":"sub"},{"color":"amber","style":"box","anchor":"suby"}],"connections":[{"to":"sub","from":"identity","color":"purple","label":"substitute","style":"dashed"}]},"order":2,"title":"Apply the Pythagorean identity","description":"We now use the fundamental Pythagorean identity $\\sin^2 t + \\cos^2 t = 1$ to link our two expressions and eliminate $t$."},{"type":"text","order":3,"title":"Solve for y","content":"We can now rearrange this equation to express $y$ in the form $f(x)$:\n\n$$\\frac{y + 5}{2} = 1 - \\frac{(x - 2)^2}{9}$$\n\nMultiply both sides by $2$:\n$$y + 5 = 2\\left(1 - \\frac{(x - 2)^2}{9}\\right)$$\n\n$$y + 5 = 2 - \\frac{2}{9}(x - 2)^2$$\n\nSubtract $5$ from both sides:\n$$y = -3 - \\frac{2}{9}(x - 2)^2$$"},{"type":"text","order":4,"title":"Determine the domain","content":"To find the domain of $f(x)$, we use the restriction on the sine function. Since $\\sin t$ is defined for all real $t$, its range is:\n$$-1 \\le \\sin t \\le 1$$\n\nSubstitute $\\sin t = \\frac{x - 2}{3}$ into the inequality:\n$$-1 \\le \\frac{x - 2}{3} \\le 1$$\n\nMultiply the entire inequality by $3$:\n$$-3 \\le x - 2 \\le 3$$\n\nAdd $2$ to all parts:\n$$-1 \\le x \\le 5$$\n\nThus, the domain for which the curve is defined is $[-1, 5]$."}]},{"id":"add0cbb1-5b08-467e-aad9-ac56e8a1b982:cartesian_conversion","label":"(b) Cartesian Equation Method","steps":[{"type":"text","order":0,"title":"Express sin(t) in terms of x","content":"To find the Cartesian equation of the curve, we first isolate the trigonometric term $\\sin t$ from the parametric equation for $x$:\n\n$$x = 3 \\sin t + 2$$\n\nSubtracting $2$ and then dividing by $3$ gives:\n\n$$\\sin t = \\frac{x - 2}{3}$$"},{"type":"text","order":1,"title":"Apply trigonometric identity to y","content":"Next, we use the fundamental trigonometric identity:\n\n$$\\cos^2 t = 1 - \\sin^2 t$$\n\nSubstituting this identity into our parametric equation for $y$:\n\n$$y = 2 \\cos^2 t - 5$$\n\n$$y = 2(1 - \\sin^2 t) - 5$$\n\n$$y = -2\\sin^2 t - 3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sin t = \\htmlData{anchor=sin-expr}{\\color{@sky}{\\frac{x - 2}{3}}}$"},{"latex":"$$y = -2 (\\htmlData{anchor=sin-sub}{\\color{@sky}{\\sin t}})^2 - 3$"},{"latex":"$$y = -\\frac{2}{9}(x - 2)^2 - 3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-expr"},{"color":"sky","style":"text-color","anchor":"sin-sub"}],"connections":[{"to":"sin-sub","from":"sin-expr","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Substitute to get Cartesian equation","description":"Now, we substitute our expression for $\\sin t$ into the simplified equation for $y$ to get the final Cartesian equation."},{"type":"text","order":3,"title":"Determine domain and endpoints","content":"Since $0 \\le t \\le 2\\pi$, the range of $\\sin t$ is restricted:\n\n$$-1 \\le \\sin t \\le 1$$\n\nWe can find the domain restriction for $x$ by substituting these limits into $x = 3 \\sin t + 2$:\n\n- Minimum value: when $\\sin t = -1$, $x = 3(-1) + 2 = -1$\n- Maximum value: when $\\sin t = 1$, $x = 3(1) + 2 = 5$\n\nSo, the domain restriction is $-1 \\le x \\le 5$.\n\nAt both endpoints $x = -1$ and $x = 5$, the corresponding $y$-value is:\n\n$$y = -\\frac{2}{9}(\\pm 3)^2 - 3 = -2 - 3 = -5$$\n\nThis gives the end coordinates of our curve: $(-1, -5)$ and $(5, -5)$."},{"type":"text","order":4,"title":"Identify the vertex","content":"The Cartesian equation is in the vertex form of a quadratic, $y = a(x-h)^2 + k$:\n\n$$y = -\\frac{2}{9}(x - 2)^2 - 3$$\n\n- Since the coefficient $-\\frac{2}{9}$ is negative, the parabola is **inverted** (opening downwards).\n- The vertex $(h, k)$ is located at $(2, -3)$, which is the maximum point of this inverted parabola."},{"type":"text","order":5,"title":"Sketch the curve","content":"Now we sketch the curve using our key features:\n\n1. Plot the vertex at $(2, -3)$ as a maximum point.\n2. Plot the endpoints at $(-1, -5)$ and $(5, -5)$.\n3. Draw a smooth, symmetric, inverted parabola shape connecting these points. Ensure your curve does not extend beyond $x = -1$ or $x = 5$.","calculator":[{"gdc":{"instructions":"1. Change your graphing mode to Parametric.\n2. Enter the equations: x(t) = 3*sin(t) + 2 and y(t) = 2*cos(t)^2 - 5.\n3. Set the domain variable t from 0 to 6.28.\n4. Set an appropriate window size to view the interval [-1, 5] on the x-axis and [-6, -2] on the y-axis."},"ti84":{"instructions":"1. Press [MODE], arrow down to 'FUNC', change it to 'PARAMETRIC' (PAR), and press [ENTER].\n2. Press [Y=] and enter:\n X1T = 3 * sin(T) + 2\n Y1T = 2 * (cos(T))^2 - 5\n3. Press [WINDOW] and set:\n Tmin = 0, Tmax = 6.28, Tstep = 0.1\n Xmin = -3, Xmax = 7, Xscl = 1\n Ymin = -7, Ymax = -1, Yscl = 1\n4. Press [GRAPH] to view the curve."},"desmos":{"expression":"(3*sin(t) + 2, 2*(cos(t))^2 - 5)","instructions":"1. Type the coordinate pair representing the parametric equations in a single line:\n (3*sin(t) + 2, 2*(cos(t))^2 - 5)\n2. Set the parameter interval beneath it to: 0 <= t <= 2pi."},"description":"Visualize the parametric curve to verify its shape, vertex, and domain endpoints."}]}]},{"id":"add0cbb1-5b08-467e-aad9-ac56e8a1b982:point_plotting","label":"(b) Point-by-Point Plotting","steps":[{"type":"text","order":0,"title":"Set up a coordinate table","content":"To sketch the parametric curve using a point-by-point plotting approach, we calculate the coordinates $(x, y)$ for key values of $t$ in the interval $0 \\le t \\le 2\\pi$.\n\nLet's choose the standard angles:\n$$t = 0, \\quad \\frac{\\pi}{2}, \\quad \\pi, \\quad \\frac{3\\pi}{2}, \\quad 2\\pi$$\n\nOur parametric equations are:\n$$x = 3 \\sin t + 2$$\n$$y = 2 \\cos^2 t - 5$$"},{"type":"text","order":1,"title":"Evaluate points for t","content":"Let's calculate the coordinates $(x, y)$ for each chosen value of $t$:\n\n1. **For $t = 0$:**\n $$x = 3 \\sin(0) + 2 = 2$$\n $$y = 2 \\cos^2(0) - 5 = 2(1)^2 - 5 = -3$$\n Point: $(2, -3)$\n\n2. **For $t = \\frac{\\pi}{2}$:**\n $$x = 3 \\sin\\left(\\frac{\\pi}{2}\\right) + 2 = 3(1) + 2 = 5$$\n $$y = 2 \\cos^2\\left(\\frac{\\pi}{2}\\right) - 5 = 2(0)^2 - 5 = -5$$\n Point: $(5, -5)$\n\n3. **For $t = \\pi$:**\n $$x = 3 \\sin(\\pi) + 2 = 2$$\n $$y = 2 \\cos^2(\\pi) - 5 = 2(-1)^2 - 5 = -3$$\n Point: $(2, -3)$\n\n4. **For $t = \\frac{3\\pi}{2}$:**\n $$x = 3 \\sin\\left(\\frac{3\\pi}{2}\\right) + 2 = 3(-1) + 2 = -1$$\n $$y = 2 \\cos^2\\left(\\frac{3\\pi}{2}\\right) - 5 = 2(0)^2 - 5 = -5$$\n Point: $(-1, -5)$\n\n5. **For $t = 2\\pi$:**\n $$x = 3 \\sin(2\\pi) + 2 = 2$$\n $$y = 2 \\cos^2(2\\pi) - 5 = 2(1)^2 - 5 = -3$$\n Point: $(2, -3)$"},{"type":"text","order":2,"title":"Summarise coordinate table","content":"Let's compile these evaluated points into a clear summary table:\n\n$$\\begin{array}{|c|c|c|c|} \\hline t & x = 3\\sin t + 2 & y = 2\\cos^2 t - 5 & (x, y) \\\\ \\hline 0 & 2 & -3 & (2, -3) \\\\ \\hline \\frac{\\pi}{2} & 5 & -5 & (5, -5) \\\\ \\hline \\pi & 2 & -3 & (2, -3) \\\\ \\hline \\frac{3\\pi}{2} & -1 & -5 & (-1, -5) \\\\ \\hline 2\\pi & 2 & -3 & (2, -3) \\\\ \\hline \\end{array}$$\n\nNotice that as $t$ goes from $0$ to $2\\pi$:\n- The curve starts at the vertex $(2, -3)$.\n- It moves to the right endpoint $(5, -5)$ at $t = \\frac{\\pi}{2}$.\n- It returns to the vertex $(2, -3)$ at $t = \\pi$.\n- It moves to the left endpoint $(-1, -5)$ at $t = \\frac{3\\pi}{2}$.\n- It returns to the vertex $(2, -3)$ at $t = 2\\pi$."},{"type":"text","order":3,"title":"Analyze the shape","content":"Let's understand the underlying algebraic shape to draw the sketch accurately.\n\nFrom our table, we can see:\n- The maximum value of $y$ is $-3$, which occurs when $x = 2$. This indicates a peak or vertex at $(2, -3)$.\n- The minimum value of $y$ is $-5$, which occurs at the boundary points $x = -1$ and $x = 5$.\n- The coordinates are symmetric about the vertical line $x = 2$.\n\nThis behavior matches an inverted (negative) parabola opening downwards, with its vertex at $(2, -3)$ and endpoints at $(-1, -5)$ and $(5, -5)$."},{"type":"text","order":4,"title":"Verify with GDC","content":"We can easily verify the plotted path and the endpoints of the curve using a graphing calculator.","calculator":[{"gdc":{"instructions":"1. Set your graphing calculator's mode to Parametric equations.\n2. Define X1(t) = 3*sin(t) + 2 and Y1(t) = 2*cos(t)^2 - 5.\n3. Adjust the interval for t to [0, 6.28].\n4. Set a suitable window view (e.g., x from -2 to 6, y from -6 to -2).\n5. Plot the graph to verify the shape."},"ti84":{"instructions":"1. Press [MODE] and change the graphing mode to PAR (parametric).\n2. Press [Y=].\n3. Set X1T = 3 * sin(T) + 2 and Y1T = 2 * (cos(T))^2 - 5.\n4. Press [WINDOW] and set Tmin = 0, Tmax = 6.28, and Tstep = 0.1.\n5. Set Xmin = -2, Xmax = 6, Ymin = -6, Ymax = -2.\n6. Press [GRAPH] to view the curve."},"desmos":{"expression":"(3*sin(t) + 2, 2*cos(t)^2 - 5)","instructions":"1. Enter the parametric coordinate pair directly into an input row: (3*sin(t) + 2, 2*cos(t)^2 - 5).\n2. Set the parameter domain bounds next to the expression as 0 <= t <= 2*pi.\n3. Adjust the graph viewport to focus on the region around x in [-2, 6] and y in [-6, -2]."},"description":"Graph the parametric equations to visualize the negative parabola shape."}]},{"type":"text","order":5,"title":"Sketch the curve","content":"Now, sketch the curve on your axes:\n\n1. Plot the vertex at $(2, -3)$ as the highest point.\n2. Plot the left endpoint at $(-1, -5)$ and the right endpoint at $(5, -5)$.\n3. Draw a smooth, symmetric, downward-opening parabolic arc connecting $(-1, -5)$ to $(2, -3)$ and then to $(5, -5)$.\n\n**Marks awarded for:**\n- **B1**: Drawing an inverted (negative) parabola shape.\n- **B1**: Showing the vertex clearly located at $(2, -3)$.\n- **B1**: Showing the curve starting/ending at $(-1, -5)$ and $(5, -5)$ with no parts of the curve drawn outside this domain."}]}],"generatedAt":"2026-06-09T03:23:21.981Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths","summaries":[{"id":"0cf3cec0-7d3e-4255-a08a-cc01a4ed3f3b","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"100c6223-1680-4a86-990e-d2ad9bf0528d","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"543fca01-78c6-49af-ac8c-db2b576575df","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"5adb3bf5-a974-44f3-b829-cd9f56fa4014","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"afd9c275-74ca-442b-82d8-681c50290b45","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":false},{"id":"cb0ab090-0a4d-4b4a-ad34-c6028469b06f","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"d6b8dd59-ebfe-49f9-b11e-a5a5e346e2b8","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"1f3b9e4e-4c6e-4a3b-a095-ff140a7d50d0","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"7b3f3471-2160-4d33-8e91-64f2aeec3242","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"7c254a8f-4c38-4870-99b5-0105397c86cf","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"96d41ac0-231d-4cf5-83f2-5ea8f306db12","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"e2db1103-cf17-4e61-8786-dbdbbf13a927","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":false},{"id":"e9d7a7f9-40a0-46fd-8404-cce418aa8297","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"1b974b05-47bd-4b1b-9b1f-3b22927a212f","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"50d7e278-f5cc-4135-ac25-412e2b9abfaf","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":false},{"id":"5a32a335-0d40-411d-87e1-2723a51994e4","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"6b9dcbff-22e8-4551-b86a-2e2781f1df7b","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"cce8998b-de31-4c16-af74-d67bb0fbe522","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"d1ae58c3-5694-4151-9072-a8c360ff032a","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false}],"title":"8.3 Curve Sketching","topicId":"13f22ac1-2516-4b0f-93f4-a700d39f70f3","topicSidebar":"$L39","topicSlug":"a-level-pure-mathematics-8-3-curve-sketching","topicTitle":"8.3 Curve Sketching"}]}],"$L3a","$L3b"]}]