A Math

A Math (IGCSE)

E Math

E Math (IGCSE)

H1 Math

H2 Math

IB AA HL Math

IB AA SL Math

IB AI HL Math

IB AI SL Math

IP Math

N(A) A Math

N(A) E Math

PSLE

PSLE Foundation Mathematics

<p>Given that&nbsp;$\mathrm{f}(x) = 3^x$, sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}(-|x|)$ on separate diagrams. Clearly indicate the intercept(s) on the graph.<span class="inline-marks">[4]</span></p>

Given that $\mathrm{f}(x) = 3^x$, sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}(-|x|)$ on separate diagrams. Clearly indicate the intercept(s) on the graph.

Sketching Graphs of $f(\lvert x\rvert)$

Transformations of Graphs

<p>Given that&nbsp;$\mathrm{f}(x) = x(x + 3)(x - 2)$, sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}(|x|)$ on separate diagrams. Clearly indicate the intercepts and asymptotes on the graph.<span class="inline-marks">[4]</span></p>

Given that $\mathrm{f}(x) = x(x + 3)(x - 2)$, sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}(|x|)$ on separate diagrams. Clearly indicate the intercepts and asymptotes on the graph.

<p>Given that&nbsp;$\mathrm{f}(x) = \dfrac{x + 3}{x - 2}$, sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}(|x|)$ on separate diagrams. Clearly indicate the intercepts and asymptotes on the graph.<span class="inline-marks">[4]</span></p>

Given that $\mathrm{f}(x) = \dfrac{x + 3}{x - 2}$, sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}(|x|)$ on separate diagrams. Clearly indicate the intercepts and asymptotes on the graph.

<p>A function is defined by $f(x)=\dfrac{3x-2}{x+4}$, where&nbsp;$x\ne -4$.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Write down the equations of the vertical and horizontal asymptotes of the graph of $y=f(x)$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">State the $x$- and $y$-intercepts of the graph of $y=f(x)$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Sketch the graph of $y=f(x)$, clearly indicating the asymptotes and intercepts.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">Hence, sketch the graph of $y=f(|x|)$ on a separate set of axes, clearly indicating any asymptotes and axial intercepts.<span class="inline-marks">[3]</span></div></li></ol>

A function is defined by $f(x)=\dfrac{3x-2}{x+4}$, where $x\ne -4$.

Sketching Rational Functions

Asymptotes and Discontinuities

Transformations of a rational function and modulus

Graphing Techniques

<p>The curve with equation $y = \mathrm{f}(x)$ for&nbsp;$x \leq b$, where&nbsp;$\mathrm{f}(x)$ is a <strong>quadratic</strong> function, intersects the $x$-axis at the points $(a,0)$ and $(b,0)$, and the $y$-axis at the point $(0,c)$. The graph of $y = \mathrm{f}(x)$ is shown in Figure 1. The scales on the $x$- and $y$-axes are the same.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1783531831399-rvzkp28l39.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODM1MzE4MzEzOTktcnZ6a3AyOGwzOS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyODM5MTA0LCJleHAiOjE3OTI4Mzk0MDR9.8W5k0a-uW7JGXSAUZtmOro23JIpSTTsXqnUKwUXSgXw" style="max-width: min(300px, 100%); height: auto; display: block;" alt="Figure 1" class="aligncenter custom-width" /></figure><div class="image-caption" style="text-align: center;">Figure 1</div></div>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><ol class="i-list"><li data-list-item-id="item-a-i" data-list-item-path="item-a/item-a-i"><div class="list-item-paragraph">Sketch the graph of $y = \mathrm{f}(\left| x \right|)$, labelling the points where the graph intersects or touches the axes.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-a-ii" data-list-item-path="item-a/item-a-ii"><div class="list-item-paragraph">Sketch the graph of $y = \frac{1}{\mathrm{f}(x)}$, labelling the points where the graph intersects or touches the axes, as well as the equations of any asymptotes.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-a-iii" data-list-item-path="item-a/item-a-iii"><div class="list-item-paragraph">Describe fully a sequence of transformations which transforms the graph of $y = \mathrm{f}(2x+1)$ onto the graph of $y = \mathrm{f}(x)$.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">The function $y = \mathrm{g}(x)$ is such that&nbsp;$\mathrm{g}(x) = \mathrm{f}(x)$ for&nbsp;$x \leq k$, and $y = \mathrm{g}^{-1}(x)$ exists.</div><ol class="i-list"><li data-list-item-id="item-b-i" data-list-item-path="item-b/item-b-i"><div class="list-item-paragraph">Find the largest possible value of $k$ in terms of $a$ and $b$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-b-ii" data-list-item-path="item-b/item-b-ii"><div class="list-item-paragraph">Using the value of $k$ found in <strong>(b)(i)</strong>, sketch the graph of $y = \mathrm{g}^{-1}(x)$ on Figure 1 in the Answer Booklet, labelling the intersections with the axes. Write down the range of $\mathrm{g}^{-1}$ in terms of $a$ and $b$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b-iii" data-list-item-path="item-b/item-b-iii"><div class="list-item-paragraph">Explain why the solution to $\mathrm{g}^{-1}(x) = x$ satisfies the equation $\mathrm{g}^{-1}(x) = \mathrm{g}(x)$.<span class="inline-marks">[1]</span></div></li></ol></li></ol>

The curve with equation $y = \mathrm{f}(x)$ for $x \leq b$, where $\mathrm{f}(x)$ is a **quadratic** function, intersects the $x$-axis at the points $(a,0)$ and $(b,0)$, and the $y$-axis at the point $(0,c)$. The graph of $y = \mathrm{f}(x)$ is shown in Figure 1. The scales on the $x$- and $y$-axes are the same.

Sketching a Function and Its Inverse

Intersections of a Function and Its Inverse

Restricting a Domain to Obtain an Inverse

Inverse Functions

Reciprocal Graph, $y=\frac{1}{f(x)}$

Graph Transformations of $f(ax+b)$

2025 ACJC P1 Q8

Functions

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Describe a sequence of $2$ transformations that transforms the curve $C$ with equation $\frac{x^2}{9} + y^2 = 1$ onto the curve $D$ with equation $(x - 2)^2 + y^2 = 1$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">The diagram shows the curve with equation $y = \mathrm{f}(x)$. The curve passes through the origin and the point $(2, 0)$ and has a minimum point at $(3, -4)$. The asymptotes of the curve are the lines $x = 1$, $y = 0$ and $y = 2$.</div><div class="graph-block" data-graph-config="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" data-width="400" data-height="250" style="width:400px;height:250px;max-width:100%;display:block;margin:0 auto;"></div><div class="list-item-paragraph">Sketch the graph of $y = \frac{1}{\mathrm{f}(x)}$, indicating the equations of any asymptotes and the coordinates of any points where it crosses the axes and of any turning points, where applicable.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">The graph of $y = \mathrm{g}(x)$ is a semi-circle with its centre at the point $(a, 0)$, where&nbsp;$a &lt; 0$. The diagram shows the graph of $y = \mathrm{g}(|x|)$, which intersects the axes at the points $(-b, 0)$, $(b, 0)$ and $(0, c)$, where&nbsp;$a \neq -b$.<span class="inline-marks">[3]</span></div><div class="graph-block" data-graph-config="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" data-width="300" data-height="200" style="width:300px;height:200px;max-width:100%;display:block;margin:0 auto;"></div><div class="list-item-paragraph">Sketch the graph of $y = \mathrm{g}(x)$, labelling the coordinates of the centre and the radius of the semi-circle, in terms of $a$ and/or $b$.</div></li></ol>

Composite Transformations

2024 MI PRU1 Q5

<p>The curve whose equation is $y = x^3 + 3$ undergoes, in succession the following transformations:</p>
<ol class="ABC-list"><li data-list-item-id="item-A" data-list-item-path="item-A"><div class="list-item-content"><span class="list-item-text">Translation by $4$ units in the negative $x$ direction.</span></div></li><li data-list-item-id="item-B" data-list-item-path="item-B"><div class="list-item-content"><span class="list-item-text">Scaling by a scale factor of $2$ parallel to the $x$-axis.</span></div></li><li data-list-item-id="item-C" data-list-item-path="item-C"><div class="list-item-content"><span class="list-item-text">Reflection about the $y$-axis.</span></div></li><li data-list-item-id="item-D" data-list-item-path="item-D"><div class="list-item-content"><span class="list-item-text">Transformation of the function $y = \mathrm{f}(|x|)$</span></div></li></ol>
<p>Find the equation of the resulting curve.</p>

The curve whose equation is $y = x^3 + 3$ undergoes, in succession the following transformations:

Translation by $4$ units in the negative $x$ direction.

Scaling by a scale factor of $2$ parallel to the $x$-axis.

Transformation of the function $y = \mathrm{f}(|x|)$

Find the equation of the resulting curve.

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">The graph of $y=\mathrm{f}(x)$ has turning points at $A(-a, 0)$, $B(a, b)$ and $C(c, -b)$ and intersects the $x$-axis at the points $D(d, 0)$ and $E(e, 0)$. The graph of $y=\mathrm{f}(x)$ is as shown in the diagram below.<span class="inline-marks">[3]</span></div></li></ol>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1769223141925-ir1uwhrzoth.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjkyMjMxNDE5MjUtaXIxdXdocnpvdGgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjgzOTEwNCwiZXhwIjoxNzkyODM5NDA0fQ.gdA1HNfnypzLSu1v2pCKLZuNkKStbfxBGjVoQSFoDNs" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>Sketch the graph of $y=\mathrm{f}(a-x)$, labelling the coordinates of the corresponding points of $A$, $B$, $C$, $D$, and $E$ clearly.<span class="inline-marks">[3]</span></p>
<ol class="abc-list" start="2"><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">The graph of $y=\mathrm{g}(x)$ has asymptotes $y=0$, $y=-m$ and $x=-n$, and intersects the $y$-axis at $P(0, p)$, where&nbsp;$p &lt; m$. The graph of $y=\mathrm{g}(x)$ is as shown in the diagram below.</div></li></ol>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1769223158996-vh24fdfkh2.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjkyMjMxNTg5OTYtdmgyNGZkZmtoMi5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyODM5MTA0LCJleHAiOjE3OTI4Mzk0MDR9.5p7-8qW3zEFqjAuBNkNAwGURlv9doVgdGfG0DIKItfE" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>Sketch on separate diagrams the graphs of</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">$y=|\mathrm{g}(x)|$ and<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$y=\mathrm{g}(|x|)$,<span class="inline-marks">[2]</span></div></li></ol>
<p>labelling clearly the asymptote(s) and coordinates of $P$ on each graph.</p>

The graph of $y=\mathrm{f}(x)$ has turning points at $A(-a, 0)$, $B(a, b)$ and $C(c, -b)$ and intersects the $x$-axis at the points $D(d, 0)$ and $E(e, 0)$. The graph of $y=\mathrm{f}(x)$ is as shown in the diagram below.

Sketch the graph of $y=\mathrm{f}(a-x)$, labelling the coordinates of the corresponding points of $A$, $B$, $C$, $D$, and $E$ clearly.

The graph of $y=\mathrm{g}(x)$ has asymptotes $y=0$, $y=-m$ and $x=-n$, and intersects the $y$-axis at $P(0, p)$, where $p < m$. The graph of $y=\mathrm{g}(x)$ is as shown in the diagram below.

Sketch on separate diagrams the graphs of

labelling clearly the asymptote(s) and coordinates of $P$ on each graph.

Graphs of Modulus Function

2025 NYJC P1 Q4

<div class="graph-block" data-graph-config="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" data-width="401" data-height="290" style="width:401px;height:290px;max-width:100%;display:block;margin:0 auto;"></div>
<p>Sketch $y = -f\left( \frac{1}{2}x \right)$</p>
<p>Sketch $y = \left| f(3 - x) \right|$</p>
<p>Sketch $y = f\left( 1 - |x| \right)$</p>

Sketch $y = -f\left( \frac{1}{2}x \right)$

<div class="graph-block" data-graph-config="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" data-width="333" data-height="237" style="width:333px;height:237px;max-width:100%;display:block;margin:0 auto;"></div>
<p>Sketch $y = 3f(2x) - 1$</p>
<p>Sketch $y = f(1 - 2x)$</p>
<p>Sketch $3y = f(-|x|)$</p>

<p>The curve whose equation is $y = e^{3x + 2} + 1$ undergoes, in succession, the following transformations:<br></br></p>
<p>A: Translation by $3$ units in the positive x direction.<br></p>
<p>B: Scaling by a scale factor of $4$ parallel to the x-axis.<br></p>
<p>C: Reflection about the $y$-axis.<br></p>
<p>D: Transformation of the function $y = f(|x|)$.<br></br></p>
<p>Find the equation of the resulting curve.</p>

The curve whose equation is $y = e^{3x + 2} + 1$ undergoes, in succession, the following transformations:<br></br>

A: Translation by $3$ units in the positive x direction.


B: Scaling by a scale factor of $4$ parallel to the x-axis.


D: Transformation of the function $y = f(|x|)$.<br></br>

<p>It is given that&nbsp;$\mathrm{f}\left( x \right)=-4{{x}^{2}}+12x+7$.</p>
<ol class="i-list"><li data-list-item-id="item-1761039907999-py32a34nh" data-list-item-path="item-1761039907999-py32a34nh"><div class="list-item-paragraph">Sketch the graph of $y=\mathrm{f}\left( \left| x \right| \right)$, stating the coordinates of turning point(s) and of point(s) where&nbsp;$y=\mathrm{f}\left( \left| x \right| \right)$ crosses the axes.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1761039908554-gmkaq9eka" data-list-item-path="item-1761039908554-gmkaq9eka"><div class="list-item-paragraph">Solve the inequality $-4{{x}^{2}}+12\left| x \right|+7\le 4x+7$.<span class="inline-marks">[2]</span></div></li></ol>

It is given that $\mathrm{f}\left( x \right)=-4{{x}^{2}}+12x+7$.

Inequalities involving Modulus Functions

2020 SAJC P1 Q1

Inequalities

<p>The function $y=\mathrm{f}\left( x \right)$ is a one-to-one function.</p>
<p>The graphs of $y=\mathrm{f}\left( \left| x \right| \right)$ and $y=\left| \mathrm{f}\left( x \right) \right|$ are shown below in Figure 1 and 2 respectively.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760435914280-9o1denki78a.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA0MzU5MTQyODAtOW8xZGVua2k3OGEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjgzOTEwNCwiZXhwIjoxNzkyODM5NDA0fQ.QQyKdyBpJQDLb0mKdscOQD5naRYQZlqGA3a7tuPwQZk" style="max-width: min(700px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-1760435897187-klomepur2" data-list-item-path="item-1760435897187-klomepur2"><div class="list-item-paragraph">State the range of values of $x$ for which $\mathrm{f}\left( x \right)\le 0$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-1760435897713-f83zntrag" data-list-item-path="item-1760435897713-f83zntrag"><div class="list-item-paragraph">State the equation(s) of the asymptote(s) of the graph of $y=\frac{1}{\mathrm{f}\left( x \right)}$.<span class="inline-marks">[2]</span></div></li></ol>

The function $y=\mathrm{f}\left( x \right)$ is a one-to-one function.

The graphs of $y=\mathrm{f}\left( \left| x \right| \right)$ and $y=\left| \mathrm{f}\left( x \right) \right|$ are shown below in Figure 1 and 2 respectively.

Graphs of $\lvert f(x)\rvert$ and $f(\lvert x\rvert)$

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<ol class="abc-list"><li><div class="list-item-paragraph">The diagram above shows the graph of $y=\mathrm{f}\left( x \right)$. The curve crosses the $x$-axis at $\left( -3.5,\,\,0 \right)$ and has a minimum point at $\left( -2,\,\,0 \right)$. The lines $x=-3$, $x=0$ and $y=3$ are asymptotes of the curve.</div><div class="list-item-paragraph">On separate diagrams, sketch the graphs in <strong>(i)</strong> and <strong>(ii)</strong>, stating the equations of any asymptotes and the coordinates of the points where the curve crosses the axes and of any turning points in exact form, if possible.</div><ol class="i-list"><li><div class="list-item-paragraph">$y=\mathrm{f}\left( \left| x \right| \right)$, and<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">$y=\mathrm{f}\,\,\mathrm{ }\!\!'\!\!\mathrm{ }\left( x \right)$.<span class="inline-marks">[4]</span></div></li></ol></li><li><div class="list-item-paragraph">The transformations A, B and C are given as follows:</div><div class="list-item-paragraph">A: $ \qquad $ Reflection about the $x$-axis;</div><div class="list-item-paragraph">B: $ \qquad $ Translation of $4$ units in the positive $x$-direction;</div><div class="list-item-paragraph">C: $ \qquad $ Scaling parallel to the $x$-axis by a factor of $2$.</div><div class="list-item-paragraph">A curve undergoes in succession, the transformations A, B and C, and the equation of the resulting curve is $y=\frac{1}{3}{{\left( x+1 \right)}^{2}}$.</div><div class="list-item-paragraph">Determine the original equation of the curve.<span class="inline-marks">[3]</span></div></li></ol>

Recovering Original Graphs from Transformations

Sketching Derivative Graphs

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Applications of Differentiation

<ol class="abc-list"><li data-list-item-id="item-1759919560416-0x54j67eb"><div class="list-item-paragraph">The diagram shows the graph of $y=\mathrm{f}\left( x \right)$. The curve passes through the origin, has a minimum point at $\left( -3,0 \right)$ and its equations of asymptotes are $x=-2$ and $y=2$.</div><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1759919573279-x5glibx4w3r.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NTk5MTk1NzMyNzkteDVnbGlieDR3M3IucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjgzOTEwNCwiZXhwIjoxNzkyODM5NDA0fQ.pRKpbG2zmXzueeF0xinBbnE_TCWhEQVtlXMGraSA8D4" alt="List item image" class="aligncenter" style="max-width: 100%; height: auto;" /></div><div class="list-item-paragraph">Sketch the following graphs on separate axes.</div><ol class="i-list"><li data-list-item-id="subitem-1759919583922-pzh5mmhqt"><div class="list-item-paragraph">$y=\frac{1}{\mathrm{f}\left( x \right)}$<span class="inline-marks">[3]</span></div></li><li data-list-item-id="subitem-1759919616181-qb3lwb6qk"><div class="list-item-paragraph">$y=\mathrm{f}\left( -\left| x \right| \right)$<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="item-1759919600654-hc3l0n9wa"><div class="list-item-paragraph">Describe a series of transformations that transforms the graph of ${{x}^{2}}+{{\left( y+1 \right)}^{2}}=1$ onto the graph of ${{\left( x-1 \right)}^{2}}+\frac{{{y}^{2}}}{9}=1$.<span class="inline-marks">[3]</span></div></li></ol>

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<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">The curve ${{C}_{1}}$ has equation $y=\mathrm{f}\left( x \right)$, where&nbsp;$\mathrm{f}\left( x \right)={{x}^{3}}-4x+1$.</span></div><div class="list-item-paragraph">A second curve ${{C}_{2}}$ undergoes, in succession, the following transformations to get the resulting curve ${{C}_{1}}$:</div><div class="list-item-paragraph" style="margin-left: 3rem">$A$: A translation of $1$ unit in the negative direction of the $y$-axis</div><div class="list-item-paragraph" style="margin-left: 3rem">$B$: A reflection in the $y$-axis</div><div class="list-item-paragraph" style="margin-left: 3rem">$C$: A scaling of factor $\frac{1}{4}$ parallel to the $x$-axis</div><div class="list-item-paragraph">Determine the equation of ${{C}_{2}}$, showing your workings clearly.</div></li><li><div class="list-item-paragraph">The diagram shows the curve $y=\mathrm{f}\left( x \right)$. The curve passes through $\left( 4,0 \right)$ and $\left( 0,-2 \right)$, and has a maximum point $\left( 6,a \right)$ where&nbsp;$a&gt;0$. It has asymptotes $x=2$, $y=\frac{1}{2}a$ and $y=0$.</div><div class="list-item-paragraph">Sketch the graph of</div><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1762971341160-kgk7n9us39.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjI5NzEzNDExNjAta2drN245dXMzOS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyODM5MTA0LCJleHAiOjE3OTI4Mzk0MDR9.37TtJlfJCaPFs6XSvqeRkyqz6MnVRoCR8-IlQMbdBdI" alt="List item image" class="aligncenter custom-width" style="max-width: min(400px, 100%); height: auto;" /></div><ol class="i-list"><li><div class="list-item-paragraph">$y=\frac{1}{\mathrm{f}\left( x \right)}$,<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">$y=\mathrm{f}\left( \left| x \right| \right)$,<span class="inline-marks">[3]</span></div><div class="list-item-paragraph">showing clearly the stationary points, axial intercepts and equations of all asymptotes.</div></li></ol></li></ol>

The curve ${{C}_{1}}$ has equation $y=\mathrm{f}\left( x \right)$, where $\mathrm{f}\left( x \right)={{x}^{3}}-4x+1$.

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<p>The diagram shows the curve $y=\mathrm{f}\left( x \right)$. There are two vertical asymptotes with equations $x=-2$ and $x=2$ respectively. The curve crosses the $x$-axis at the point $A$ and has a maximum turning point at $B$ where it crosses the $y$-axis.</p>
<p>The curve also has a minimum turning point at $C$. The coordinates of $A$, $B$ and $C$ are $\left( a,0 \right)$, $\left( 0,-10 \right)$ and $\left( p,q \right)$ respectively, where&nbsp;$a$, $p$ and $q$ are constants</p>
<p>Sketch the following curves and state the equations of the asymptotes, the coordinates of the turning points and of points where the curve crosses the axes, if any. Leave your answers in terms of $a$, $p$ or $q$ where necessary.</p>
<ol class="i-list"><li data-list-item-id="item-1759226109016-pkmxk2gk7" data-list-item-path="item-1759226109016-pkmxk2gk7"><div class="list-item-paragraph">$y=\frac{1}{\mathrm{f}\left( x \right)}$, and<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1759226109622-kqp84utzn" data-list-item-path="item-1759226109622-kqp84utzn"><div class="list-item-paragraph">$y=\mathrm{f}\left( 2-\left| x \right| \right)$. .<span class="inline-marks">[3]</span></div></li></ol>

The diagram shows the curve $y=\mathrm{f}\left( x \right)$. There are two vertical asymptotes with equations $x=-2$ and $x=2$ respectively. The curve crosses the $x$-axis at the point $A$ and has a maximum turning point at $B$ where it crosses the $y$-axis.

The curve also has a minimum turning point at $C$. The coordinates of $A$, $B$ and $C$ are $\left( a,0 \right)$, $\left( 0,-10 \right)$ and $\left( p,q \right)$ respectively, where $a$, $p$ and $q$ are constants

Sketch the following curves and state the equations of the asymptotes, the coordinates of the turning points and of points where the curve crosses the axes, if any. Leave your answers in terms of $a$, $p$ or $q$ where necessary.

$y=\frac{1}{\mathrm{f}\left( x \right)}$, and

$y=\mathrm{f}\left( 2-\left| x \right| \right)$. .

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<ol class="abc-list"><li><div class="list-item-paragraph">Describe a sequence of three transformations which transforms the graph of $y=x-\frac{1}{2x+4}$ to the graph of $y=-2\left( x+2 \right)+\frac{1}{x}$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">The diagram shows the graph of $y=\mathrm{f}\left( x \right)$. The curve passes through the points $\left( -3,-2 \right)$, $\left( -1,0 \right)$, $\left( 0,1 \right)$, $\left( 1,2 \right)$ and $\left( 3,0 \right)$, and has a horizontal asymptote at $y=-\frac{1}{2}$.</div><div class="list-item-paragraph">On separate diagrams that are clearly labelled, sketch the graphs of</div><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1764583914468-zia0m90by9.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjQ1ODM5MTQ0NjgtemlhMG05MGJ5OS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyODM5MTA0LCJleHAiOjE3OTI4Mzk0MDR9.FN5k2CmwFduwzP7vQU3VwfXJPc8NZx__vxpS770Cl3s" alt="List item image" class="aligncenter custom-width" style="max-width: min(450px, 100%); height: auto;" /></div><ol class="i-list"><li><div class="list-item-paragraph">$y=\frac{1}{\mathrm{f}\left( x \right)}$,<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">$y=\mathrm{f}\left( \left| x \right|-1 \right)$,<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">$y=\mathrm{f}\left( 2x+1 \right)$.<span class="inline-marks">[3]</span></div></li></ol></li></ol>

Describe a sequence of three transformations which transforms the graph of $y=x-\frac{1}{2x+4}$ to the graph of $y=-2\left( x+2 \right)+\frac{1}{x}$.

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