Read a derivative and reciprocal from a shifted graph · source reference

Printed question

(a) A curve $y=f(x)$ is transformed by a stretch with scale factor 2 parallel to the $y$-axis, followed by a reflection about the $x$-axis. The equation of the resultant curve is $y=\ln[\frac1{(x+a)^2}]$, where $x>-a$ and $a>1$. Find $f(x)$ and sketch the graph of $y=f(x)$. [4]

(b) The diagram shows the graph of $y=g(x+1)$. The graph passes through the origin and has a turning point $A(3,4)$. The asymptotes of the graph are $x=2$ and $y=2x-4$.

On separate diagram, sketch the following graphs indicating the points corresponding to the axial intercepts, turning point and asymptotes, where necessary.

(i) $y=g(x)$ [2]

(ii) $y=g'(x)$ [3]

(iii) $y=\frac1{g(x)}$ [3]

[modified 2015/RVHS/Prelim/I/10]

Printed working

(a) $y=f(x)\xrightarrow A y=2f(x)\xrightarrow B y=-2f(x)$, where $A$: scale with factor 2 parallel to the $y$-axis; $B$: reflection about the $x$-axis.

Thus $-2f(x)=\ln[\frac1{(x+a)^2}]$.

$\Rightarrow-2f(x)=\ln(x+a)^{-2}$.

$\Rightarrow-2f(x)=-2\ln(x+a)$.

Therefore, $f(x)=\ln(x+a)$.

(b)(i) $y=g(x+1)\xrightarrow{x\mapsto x-1}y=g(x)$.

(ii) $y=g'(x)$.

(iii) $y=\frac1{g(x)}$.

Teacher notes

(a) A: Replace $y$ by $\frac12y$. B: Replace $y$ by $-y$.

(b)(i) Translate 1 unit in positive $x$-direction.

(b)(ii) 1. Vertical asymptote $x=3$ remains. Oblique asymptote $y=2x-6$ of curve $y=g(x)$ becomes horizontal asymptote $y=2$ of curve $y=g'(x)$.

2. Turning pt $A'(4,4)$ of curve $y=g(x)$ becomes $x$-intercept $(4,0)$ of curve $y=g'(x)$ as $g'(x)=0$ at stationary pt.

3. Map out the curve $y=g'(x)$ by tracing the gradient of curve $y=g(x)$ as $x$ increases. As $x$ increases from $-\infty$ to 3, gradient increases from $2^+$ to $\infty$; as $x$ increases from 3 to 4, gradient increases from $-\infty$ to 0; as $x$ increases from 4 to $\infty$, gradient increases from 0 to $2^-$.

(b)(iii) 1. Divide original graph into regions where $y>0,y<0$. Note $g(x)$ and $\frac1{g(x)}$ have the same sign.

2. Take the reciprocal of all $y$-coords on graph of $y=g(x)$ (i.e. $\frac1{g(x)}$).

3. Identify new intercepts, asymptotes and turning pts.

4. Within each region, complete the graph using the features identified in 2,3.

Printed comparison table

$y=g(x)$	$y=1/g(x)$
Vertical asymptote $x=3$	$x$-intercept $(3,0)$
Oblique asymptote $y=2x-6$	Horizontal asymptote $y=0$
$x$-intercept $(1,0)$	Vertical asymptote $x=1$
Min pt $(4,4)$	Max pt $(4,1/4)$
As $x$ increases, $y$ increases; as $x$ increases, $y$ decreases	As $x$ increases, $y$ decreases; as $x$ increases, $y$ increases
$y>0$ ($y<0$)	$y>0$ ($y<0$)

Printed answer key

The answer key repeats the native diagrams above with the same coordinates and asymptote labels; no additional algebraic answer is printed.

Editorial qualification

The printed reciprocal table calls (3,0) an x-intercept. It is a limiting hole because g is undefined at 3. Differentiation sketch follows the specific source shape; the stated asymptotes alone would not imply derivative asymptotes for every arbitrary differentiable function.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: GRAPHS AND CURVE SKETCHING. Question/answer booklet printed pages 1–14; solutions printed pages 1–28. Running metadata and repeated headings are transcribed in the topic metadata article.