Read stationary points and slope limits · source reference

Printed question

The diagram below shows the graph of $y=f(x)$. The curve cuts the axes at $(-1,0)$ and $(0,3)$. It has a maximum point at $(3,5)$, and asymptotes $x=-2$ and $y=2$.

Sketch the graph of $y=f'(x)$, showing clearly asymptotes, intercepts and coordinates of turning points, where available. [3]

[modified 2010/ACJC/Prelim/I/3]

Printed working

The derivative sketch has vertical asymptote $x=-2$, horizontal asymptote $y=0$, intercept $(3,0)$, positive branches before $x=3$ (excluding $-2$), and a negative trough after $3$ tending to $0$ from below. No exact coordinates of that trough are supplied.

Teacher notes

1. Vertical asymptote $x=-2$ remains. Horizontal asymptote $y=2$ of curve $y=f(x)$ becomes horizontal asymptote $y=0$ of curve $y=f'(x)$ since, as curve $y=f(x)$ approaches the horizontal asymptote, its gradient gradually approaches $0$. The source shorthand is “$f(x)=2\Rightarrow f'(x)=0$”; this means the asymptotic gradient in the drawn curve, not equality of $f$ with $2$ everywhere.

2. Stationary point $(3,5)$ of curve $y=f(x)$ becomes $x$-intercept $(3,0)$ of curve $y=f'(x)$ as $f'(x)=0$ at a stationary point.

3. Map out the curve $y=f'(x)$ by tracing the gradient of curve $y=f(x)$ as $x$ increases.

Answer key

Same derivative sketch as the detailed solution.

Native diagrams