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PSLE

PSLE Foundation Mathematics

<p>A curve $C$ has equation $y=\dfrac{x^2-4x+1}{x+1}$, $x \ne -1$.</p>
<ol class="abc-list"><li data-list-item-id="h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-0" data-list-item-path="h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-0"><div class="list-item-paragraph">Prove, using an algebraic method, that&nbsp;$y=\dfrac{x^2-4x+1}{x+1}$ cannot lie between two certain values, to be determined in exact form.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-1" data-list-item-path="h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-1"><div class="list-item-paragraph">Sketch $C$, stating the equations of any asymptotes, and the coordinates of any turning points and any axial intercepts.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-2" data-list-item-path="h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-2"><div class="list-item-paragraph">The curve $C$ undergoes, in succession, the following transformations.</div><ol class="num-list"><li data-list-item-id="h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-2-sub-0" data-list-item-path="h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-2/h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-2-sub-0"><div class="list-item-paragraph">A translation of 1 unit in the positive $x$-direction,</div></li><li data-list-item-id="h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-2-sub-1" data-list-item-path="h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-2/h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-2-sub-1"><div class="list-item-paragraph"><p>A stretch parallel to the $x$-axis, factor $3$, $y$-axis invariant,</p></div></li><li data-list-item-id="h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-2-sub-2" data-list-item-path="h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-2/h2-2025-vjc-promo-q9-vjc-q9-item-block-1-item-2-sub-2"><div class="list-item-paragraph">A reflection in the $y$-axis.</div></li></ol><div class="list-item-after-content">Find the equation of the new curve in the form $y=\mathrm{f}(x)$.<span class="inline-marks">[3]</span></div></li></ol>

A curve $C$ has equation $y=\dfrac{x^2-4x+1}{x+1}$, $x \ne -1$.

Prove, using an algebraic method, that $y=\dfrac{x^2-4x+1}{x+1}$ cannot lie between two certain values, to be determined in exact form.

Sketch $C$, stating the equations of any asymptotes, and the coordinates of any turning points and any axial intercepts.

Composite Transformations

Sketching Rational Functions

Asymptotes and Discontinuities

Finding Possible y-values of a Curve

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Graphing Techniques

Transformations of Graphs

<ol class="abc-list"><li data-list-item-id="h2-2025-mi-promo-q7-mi-q7-item-block-0-item-0" data-list-item-path="h2-2025-mi-promo-q7-mi-q7-item-block-0-item-0"><div class="list-item-paragraph">Given that&nbsp;$a$ is a positive constant, describe a sequence of transformations which transform the graph of $y=e^x$ to the graph of $y=ae^{-x+6}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-2025-mi-promo-q7-mi-q7-item-block-0-item-1" data-list-item-path="h2-2025-mi-promo-q7-mi-q7-item-block-0-item-1"><div class="list-item-paragraph"><p>The function g and h are defined by</p><p>\[\begin{align*} \mathrm{g} &amp;: x \mapsto ae^{-x+6}, \quad x \in \mathbb{R}, \quad a &gt; 0, \\ \mathrm{h} &amp;: x \mapsto 6 + \ln x, \quad x &gt; 0 \end{align*}\]</p></div><ol class="i-list"><li data-list-item-id="h2-2025-mi-promo-q7-mi-q7-item-block-0-item-1-sub-0" data-list-item-path="h2-2025-mi-promo-q7-mi-q7-item-block-0-item-1/h2-2025-mi-promo-q7-mi-q7-item-block-0-item-1-sub-0"><div class="list-item-paragraph"><p>Find the range of $g$.</p><span class="inline-marks">[1]</span></div></li><li data-list-item-id="h2-2025-mi-promo-q7-mi-q7-item-block-0-item-1-sub-1" data-list-item-path="h2-2025-mi-promo-q7-mi-q7-item-block-0-item-1/h2-2025-mi-promo-q7-mi-q7-item-block-0-item-1-sub-1"><div class="list-item-paragraph">Hence determine whether the function hg exists.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="h2-2025-mi-promo-q7-mi-q7-item-block-0-item-1-sub-2" data-list-item-path="h2-2025-mi-promo-q7-mi-q7-item-block-0-item-1/h2-2025-mi-promo-q7-mi-q7-item-block-0-item-1-sub-2"><div class="list-item-paragraph"><p>Given instead that&nbsp;$a=3$.</p></div><ol class="num-list"><li data-list-item-id="item-1789110199479-w9h9bsfum" data-list-item-path="item-1789110199479-w9h9bsfum"><div class="list-item-paragraph"><p>Find $\mathrm{hg}(x)$ and state the domain of $\mathrm{hg}$.</p><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1789110210271-d4buk4zf3" data-list-item-path="item-1789110210271-d4buk4zf3"><div class="list-item-paragraph"><p>Find the exact range of $\mathrm{hg}$.</p><span class="inline-marks">[2]</span></div></li></ol></li></ol></li></ol>

Given that $a$ is a positive constant, describe a sequence of transformations which transform the graph of $y=e^x$ to the graph of $y=ae^{-x+6}$.

Composite Functions

Existence of Composite Functions

Range of a Composite Function

Range of a Function

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Functions

<p>The curve whose equation is $y=3-e^x$ undergoes, in succession, the following transformations:</p>
<p>A: Translation by $2$ units in the positive $x$-direction.</p>
<p>B: Scaling by a scale factor of $\frac{1}{2}$ parallel to the $x$-axis.</p>
<p>C: Translation by $1$ unit in the positive $y$-direction.</p>
<p>D: Scaling by a scale factor of $2$ parallel to the $y$-axis.</p>
<p>Find the equation of the resulting curve.<span class="inline-marks">[5]</span></p>

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

A: Translation by $2$ units in the positive $x$-direction.

B: Scaling by a scale factor of $\frac{1}{2}$ parallel to the $x$-axis.

C: Translation by $1$ unit in the positive $y$-direction.

D: Scaling by a scale factor of $2$ parallel to the $y$-axis.

Find the equation of the resulting curve.

Successive transformations of an exponential curve

<p>The curve whose equation is $y=x^3+2$ undergoes, in succession, the following transformations:</p>
<p>A: Translation by $3$ units in the positive $x$-direction.</p>
<p>B: Scaling by a scale factor of $2$ parallel to the $x$-axis.</p>
<p>C: Translation by $4$ units in the negative $y$-direction.</p>
<p>D: Scaling by a scale factor of $3$ parallel to the $y$-axis.</p>
<p>Find the equation of the resulting curve.<span class="inline-marks">[5]</span></p>

The curve whose equation is $y=x^3+2$ undergoes, in succession, the following transformations:

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

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

C: Translation by $4$ units in the negative $y$-direction.

D: Scaling by a scale factor of $3$ parallel to the $y$-axis.

Successive transformations of a cubic curve

<p>The curve whose equation is ${y = \ln(2-x)}$ undergoes, in succession the following transformations:<br>A: Translation by ${5}$ units in the positive ${x}$-direction.<br>B: Scaling by a scale factor of ${\frac{1}{3}}$ parallel to the ${x}$-axis.<br>C: Translation of ${5}$ units in the negative ${y}$-direction.<br>D: Scaling by a scale factor of ${\frac{1}{2}}$ parallel to the ${y}$-axis.<br>Find the equation of the resulting curve.</p>

The curve whose equation is ${y = \ln(2-x)}$ undergoes, in succession the following transformations:
A: Translation by ${5}$ units in the positive ${x}$-direction.
B: Scaling by a scale factor of ${\frac{1}{3}}$ parallel to the ${x}$-axis.
C: Translation of ${5}$ units in the negative ${y}$-direction.
D: Scaling by a scale factor of ${\frac{1}{2}}$ parallel to the ${y}$-axis.
Find the equation of the resulting curve.

<p>The curve whose equation is ${y = \frac{1}{x-1}}$ undergoes, in succession the following transformations:<br>A: Translation by ${2}$ units in the negative ${x}$ direction.<br>B: Scaling by a scale factor of ${3}$ parallel to the ${x}$-axis.<br>C: Translation of ${6}$ units in the positive ${y}$-direction.<br>D: Scaling by a scale factor of ${5}$ parallel to the ${y}$-axis.<br>Find the equation of the resulting curve.</p>

The curve whose equation is ${y = \frac{1}{x-1}}$ undergoes, in succession the following transformations:
A: Translation by ${2}$ units in the negative ${x}$ direction.
B: Scaling by a scale factor of ${3}$ parallel to the ${x}$-axis.
C: Translation of ${6}$ units in the positive ${y}$-direction.
D: Scaling by a scale factor of ${5}$ parallel to the ${y}$-axis.
Find the equation of the resulting curve.

<p>The diagram shows the graph of $y = \mathrm{f}(x)$ with asymptotes $x = 0$ and $y = 2x - 1$. The graph has a turning point at $\left(2, \frac{21}{4}\right)$ and it cuts the $x$-axis at $x = a$ where&nbsp;$a &lt; 0$.</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/1779465259176-jks9ys3gxbb.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Nzk0NjUyNTkxNzYtamtzOXlzM2d4YmIucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjgzOTI3NiwiZXhwIjoxNzkyODM5NTc2fQ.Y3Y5MaHcoN4qKLgsGk3_tuBq8kdwJ5gaxkbpEZxsbwg" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Sketch the graph of $y = \mathrm{f}'(x)$, stating the equations of any asymptotes and the coordinates of any turning points and points of intersection with the axes.<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">The curve $y = \mathrm{f}(x)$ undergoes the transformations $A$, $B$ and $C$ in succession.<br><br>$\qquad A$: A translation of $2$ units in the negative $x$-direction.<br>$\qquad B$: A stretch parallel to the $y$-axis with scale factor $2$.<br>$\qquad C$: A translation of $1$ unit in the positive $y$-direction.<br><br>Find the equations of the asymptotes of the resulting curve.<span class="inline-marks">[3]</span></div></li></ol>

The diagram shows the graph of $y = \mathrm{f}(x)$ with asymptotes $x = 0$ and $y = 2x - 1$. The graph has a turning point at $\left(2, \frac{21}{4}\right)$ and it cuts the $x$-axis at $x = a$ where $a < 0$.

Sketching Derivative Graphs

2024 MI PRU2 P1 Q2

Applications of Differentiation

<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>

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

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

2024 MI PRU1 Q5

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">The diagram below shows the graph of $y = \mathrm{f}(x)$. The curve passes through the points $A(-a,0)$ and $B(0,-a)$ and has a minimum point at $C(2a,-2a)$, where&nbsp;$a &gt; 0$. The lines $x = -2a$ and $y = -a$ are asymptotes of the curve.</div><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1779453402163-dw91vlxqcgf.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Nzk0NTM0MDIxNjMtZHc5MXZseHFjZ2YucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjgzOTI3NiwiZXhwIjoxNzkyODM5NTc2fQ.Qc3zsI--TKG7o8I4vzIwNzA9zRAcOGu3LxA5X31vdRw" alt="List item image" class="aligncenter custom-width" style="max-width: min(350px, 100%); height: auto;" /></div><div class="list-item-paragraph">On separate diagrams, sketch the graph of </div><ol class="i-list"><li data-list-item-id="subitem-i" data-list-item-path="item-a/subitem-i"><div class="list-item-paragraph">$y = \mathrm{f}(x-a) + a$,<span class="inline-marks">[3]</span></div></li><li data-list-item-id="subitem-ii" data-list-item-path="item-a/subitem-ii"><div class="list-item-paragraph">$y = \frac{1}{\mathrm{f}(x)}$,<span class="inline-marks">[3]</span></div></li></ol></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">The curve whose equation is $y = \sin x$, undergoes in succession, the following transformations:<br><br>$\mathbf{A}$: A translation by $\pi$ units in the negative $x$-direction.<br>$\mathbf{B}$: A scaling parallel to the $x$-axis by a factor of $3$.<br>$\mathbf{C}$: A translation by $2$ units in the positive $y$-direction.<br><br>Find the equation of the resulting curve.<span class="inline-marks">[3]</span></div></li></ol>

2024 JPJC Promo Q8

<p>It is given that&nbsp;$\mathrm{f}\left( x \right)={{\left( 1-2x \right)}^{3}}-2$ is the final graph after $\mathrm{g}\left( x \right)={{x}^{3}}$ undergoes exactly $4$ transformations. Describe the transformations involved and state the coordinates in which $\left( 0,0 \right)$ is mapped onto the graph of $\mathrm{f}\left( x \right)$.</p>

It is given that $\mathrm{f}\left( x \right)={{\left( 1-2x \right)}^{3}}-2$ is the final graph after $\mathrm{g}\left( x \right)={{x}^{3}}$ undergoes exactly $4$ transformations. Describe the transformations involved and state the coordinates in which $\left( 0,0 \right)$ is mapped onto the graph of $\mathrm{f}\left( x \right)$.

<ol class="abc-list"><li data-list-item-id="item-1769597605016-2gv61rxm1"><div class="list-item-paragraph">A curve with equation $y=\mathrm{f}(x)$ undergoes the following sequence of transformation.</div><div class="list-item-paragraph">$\qquad A$:	A translation of $3$ units in the positive $y$-direction.<br>$\qquad B$:	A translation of $2$ units in the positive $x$-direction.<br>$\qquad C$: 	A stretch with scale factor $2$ parallel to the $x$-axis.</div><div class="list-item-paragraph">The resulting curve has equation $y=\mathrm{g}(x)$ where&nbsp;$\mathrm{g}(x)=2-\frac{4}{3x-7}$. Find $\mathrm{f}(x)$ in terms of $x$.  		<span class="inline-marks">[3]</span></div><div class="list-item-paragraph">Another curve with equation $y=\frac{1}{\mathrm{f}(x)}$ undergoes the same sequence of transformations $A$, $B$ and $C$. <br>Will the equation of the resulting curve be$y=\frac{1}{\mathrm{g}(x)}$? Justify your answer<span class="inline-marks">[1]</span></div></li></ol>

Recovering Original Graphs from Transformations

2015 IJC P1 Q8

<p>The curve $C$ has equation $4{{x}^{2}}+{{y}^{2}}+8mx-4y+4=0$ , where&nbsp;$m\in \mathbb{R},m&gt;0$.<br>Show that&nbsp;$C$ can be expressed in the form  ${{\left( \frac{x+a}{b} \right)}^{2}}+{{\left( \frac{y+c}{d} \right)}^{2}}=1$, where&nbsp;$a$, $b$, $c$ and $d$ are constants to be determined in terms of $m$. <br>Hence describe a sequence of transformations, in terms of $m$, which transforms the graph of ${{x}^{2}}+{{\left( y-\frac{1}{m} \right)}^{2}}=1$ to $C$. </p>

The curve $C$ has equation $4{{x}^{2}}+{{y}^{2}}+8mx-4y+4=0$ , where $m\in \mathbb{R},m>0$.
Show that $C$ can be expressed in the form  ${{\left( \frac{x+a}{b} \right)}^{2}}+{{\left( \frac{y+c}{d} \right)}^{2}}=1$, where $a$, $b$, $c$ and $d$ are constants to be determined in terms of $m$. 
Hence describe a sequence of transformations, in terms of $m$, which transforms the graph of ${{x}^{2}}+{{\left( y-\frac{1}{m} \right)}^{2}}=1$ to $C$. 

Graphs of Conic Sections

2015 HCI Promo Q12

<p>A graph with equation $y=\mathrm{f}\left( x \right)$ undergoes $3$ successive transformations as follows:<br><br>I:	A scaling parallel to $x$-axis with scale factor $\frac{1}{2}$,<br>II:	A translation of $3$ units in the direction of the positive $x$-axis, <br>III:	A scaling parallel to $y$-axis with scale factor $2$.<br><br>The resulting equation is $y=2\ln \left( 2x-6 \right)$. Determine the equation of the graph $y=\mathrm{f}\left( x \right)$, showing your workings clearly.<br></p>

A graph with equation $y=\mathrm{f}\left( x \right)$ undergoes $3$ successive transformations as follows:

I:	A scaling parallel to $x$-axis with scale factor $\frac{1}{2}$,
II:	A translation of $3$ units in the direction of the positive $x$-axis, 
III:	A scaling parallel to $y$-axis with scale factor $2$.

The resulting equation is $y=2\ln \left( 2x-6 \right)$. Determine the equation of the graph $y=\mathrm{f}\left( x \right)$, showing your workings clearly.


2006 CJC Promo Q3 [Modified]

<p>Describe the sequence of $3$ transformations by which the graph ${{\left( 3-x \right)}^{2}}+\frac{1}{{{\left( 1-2y \right)}^{2}}}=1$ may be obtained from the graph of ${{x}^{2}}+\frac{1}{{{y}^{2}}}=1$.</p>

Describe the sequence of $3$ transformations by which the graph ${{\left( 3-x \right)}^{2}}+\frac{1}{{{\left( 1-2y \right)}^{2}}}=1$ may be obtained from the graph of ${{x}^{2}}+\frac{1}{{{y}^{2}}}=1$.

<p>Describe the sequence of transformations by which the graph ${{\left( 1-2x \right)}^{2}}+{{\left( 3-y \right)}^{2}}=1$ may be obtained from the graph of ${{x}^{2}}+{{y}^{2}}=1$.</p>

Describe the sequence of transformations by which the graph ${{\left( 1-2x \right)}^{2}}+{{\left( 3-y \right)}^{2}}=1$ may be obtained from the graph of ${{x}^{2}}+{{y}^{2}}=1$.

<p> Describe the sequence of transformations by which the graph ${{\left( \frac{x}{4}+2 \right)}^{2}}-{{\left( 5y-3 \right)}^{2}}=1$may be obtained from the graph of ${{x}^{2}}-{{y}^{2}}=1$.</p>

 Describe the sequence of transformations by which the graph ${{\left( \frac{x}{4}+2 \right)}^{2}}-{{\left( 5y-3 \right)}^{2}}=1$may be obtained from the graph of ${{x}^{2}}-{{y}^{2}}=1$.

<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 $y = \mathrm{f}(|x|)$

<p>The diagram below shows a sketch of the graph of $y = \mathrm{f}(x)$. The curve passes through the points with coordinates $(1, 0)$ and $(3, 0)$, and has turning points at $(1, 0)$ and $(4, 4)$. The asymptotes are $x = 0$, $x = 2$ and $y = 2$.</p>
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<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-content"><span class="list-item-text">Sketch on separate diagrams, the graphs of $y = \mathrm{f}'(x)$,</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-a-ii" data-list-item-path="item-a/item-a-ii"><div class="list-item-content"><span class="list-item-text">Sketch on separate diagrams, the graphs of $y = \frac{1}{\mathrm{f}(x)}$,</span><span class="inline-marks">[3]</span></div></li></ol></li></ol>
<p>showing clearly the main features of the graphs.</p>
<ol class="abc-list" start="2"><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-content"><span class="list-item-text">Describe a sequence of transformations which transform the graph of $y = \frac{x^2 + 1}{x}$ to the graph of $y = \frac{2x^2 - 5x + 4}{x - 2}$.</span><span class="inline-marks">[3]</span></div></li></ol>

The diagram below shows a sketch of the graph of $y = \mathrm{f}(x)$. The curve passes through the points with coordinates $(1, 0)$ and $(3, 0)$, and has turning points at $(1, 0)$ and $(4, 4)$. The asymptotes are $x = 0$, $x = 2$ and $y = 2$.

showing clearly the main features of the graphs.

Describe a sequence of transformations which transform the graph of $y = \frac{x^2 + 1}{x}$ to the graph of $y = \frac{2x^2 - 5x + 4}{x - 2}$.

2025 RI P1 Q5

<p>A curve has equation $y = \mathrm{f}(x)$, where&nbsp;$\mathrm{f}(x) = 1 - \sqrt{q^2 - x^2}$ for&nbsp;$q &gt; 1$. State the shape of $y = \mathrm{f}(x)$.<span class="inline-marks">[1]</span></p>
<ol class="i-list"><li data-list-item-id="item-3-i" data-list-item-path="item-3-i"><div class="list-item-paragraph">Sketch the curve $y = \frac{1}{\mathrm{f}(x)}$, giving the equations of any asymptotes and the coordinates of the end-points.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-3-ii" data-list-item-path="item-3-ii"><div class="list-item-paragraph">Describe the transformations that map the graph of $y = \mathrm{f}(x)$ to $y = -\sqrt{1-x^2}$.<span class="inline-marks">[3]</span></div></li></ol>

A curve has equation $y = \mathrm{f}(x)$, where $\mathrm{f}(x) = 1 - \sqrt{q^2 - x^2}$ for $q > 1$. State the shape of $y = \mathrm{f}(x)$.

Sketch the curve $y = \frac{1}{\mathrm{f}(x)}$, giving the equations of any asymptotes and the coordinates of the end-points.

Describe the transformations that map the graph of $y = \mathrm{f}(x)$ to $y = -\sqrt{1-x^2}$.

2025 NJC P1 Q4

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Show that&nbsp;$y = \ln \left( \frac{e^2}{3x} \right)$ can be written in the form $y = a + b \ln (cx)$, where&nbsp;$a$, $b$ and $c$ are integers to be found. Hence, state a sequence of transformations which transform the graph of $y = \ln x$ onto the graph of $y = \ln \left( \frac{e^2}{3x} \right)$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">The curve $y = \mathrm{f}(x)$ passes through the point $P$ with coordinates $(a, b)$, where&nbsp;$b \neq 0$. The tangent to the curve at $P$ has gradient $5$. When&nbsp;$y = \mathrm{f}(x)$ is transformed onto the curve $y = \mathrm{g}(x)$, $P$ corresponds to the point $R$ on $y = \mathrm{g}(x)$. For each of the following curves, state the coordinates of $R$ and find the gradient of the curve at $R$.</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">$\mathrm{g}(x) = 2\mathrm{f}(x - 1)$<span class="inline-marks">[3]</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">$\mathrm{g}(x) = \frac{1}{\mathrm{f}(x)}$<span class="inline-marks">[2]</span></div></li></ol></li></ol>

Show that $y = \ln \left( \frac{e^2}{3x} \right)$ can be written in the form $y = a + b \ln (cx)$, where $a$, $b$ and $c$ are integers to be found. Hence, state a sequence of transformations which transform the graph of $y = \ln x$ onto the graph of $y = \ln \left( \frac{e^2}{3x} \right)$.

The curve $y = \mathrm{f}(x)$ passes through the point $P$ with coordinates $(a, b)$, where $b \neq 0$. The tangent to the curve at $P$ has gradient $5$. When $y = \mathrm{f}(x)$ is transformed onto the curve $y = \mathrm{g}(x)$, $P$ corresponds to the point $R$ on $y = \mathrm{g}(x)$. For each of the following curves, state the coordinates of $R$ and find the gradient of the curve at $R$.

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