Turn a curve sketch into its gradient graph · source reference

Printed question

The diagram shows the graph of $y=f(x)$. The curve passes through the origin $O$, the points $A(-4a,0)$ and $C(-2a,0)$. The points $B(-3a,2)$ and $D(-a,-2)$ are stationary points and the lines $y=0$ and $x=a$ are asymptotes of the curve.

Sketch the graph of $y=f'(x)$, indicating clearly the points corresponding to $A$, $B$, $C$ and $D$, and the asymptotes, where applicable. [2]

[modified 2010/RVHS/Prelim/II/1]

Printed working

The printed solution is the native derivative sketch below: $B'(-3a,0)$ and $D'(-a,0)$ are its intercepts, with asymptotes $x=a$ and $y=0$. At the abscissae of $A$ and $C$, the derivative is respectively positive and negative; numeric ordinates are not supplied.

Teacher notes

1. Asymptotes: asymptotes $x=a$, $y=0$ of curve $y=f(x)$ remain as asymptotes of curve $y=f'(x)$ in the supplied sketch.

2. Stationary points $B,D$ of curve $y=f(x)$ become $x$-intercepts $B',D'$ of curve $y=f'(x)$ since $f'(x)=0$ at $B,D$.

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

Note: curve $y=f(x)$ decreases (i.e. curve $B$ to $D$) / increases ⇒ gradient $<0$ / $>0$ ⇒ curve $y=f'(x)$ lies below / above the $x$-axis.

Qualification: the asymptote rule describes this particular source sketch, not every differentiable function with a horizontal asymptote.

Answer key

Same derivative sketch as the detailed solution.

Native diagrams