Recover a rational curve from its asymptotes · source reference

Printed question

The diagram above shows the graph of the function $y=f(x)$ where $f(x)=\frac{ax^2+bx+c+1}{x+d}$, $a,b,c,d\in\mathbb R$. The oblique asymptote intersects the axes at $x=\frac32$ and $y=3$. Find the values of $a,b,d$.

By considering $\frac{dy}{dx}$, show that $c>-1$. [5]

Sketch, on a separate diagram, the graph of $y=f'(x)$. [3]

[2008/NJC/Prelim/I/11(i)]

Printed working

From the diagram vertical asymptote is $x=0$. Thus $d=0$.

Gradient of oblique asymptote $=\frac{-3}{3/2}=-2$. Equation of oblique asymptote: $y=-2x+3$.

$$y=-2x+3+\frac p x=\frac{-2x^2+3x+p}{x}.$$

Comparing coefficients, $a=-2$, $b=3$, $p=c+1$.

$$y=-2x+3+\frac{c+1}{x},\qquad \frac{dy}{dx}=-2-\frac{c+1}{x^2}.$$

From graph, $C$ has no stationary points. So $\frac{dy}{dx}=-2-\frac{c+1}{x^2}=0$ has no real solutions ⇒ $\frac{c+1}{x^2}=-2$ has no real solutions ⇒ $c+1>0$ ⇒ $c>-1$.

The source argument also uses the visible vertical asymptote: $c+1=0$ would remove it, so equality is excluded. The native derivative sketch is below.

Teacher notes

To find a vertical asymptote put denominator $=0$: $x+d=0\Rightarrow x=-d$ (provided the numerator does not cancel that pole).

Alternative no-real-solutions test: $\frac{c+1}{x^2}=-2\Rightarrow x^2=\frac{c+1}{-2}$. Since $x^2\ge0$ for real $x$, for no real solutions $\frac{c+1}{-2}<0\Rightarrow c+1>0$.

OR: $\frac{c+1}{x^2}=-2\Rightarrow2x^2+(c+1)=0$. For no real solutions the discriminant $-4(2)(c+1)<0\Rightarrow c+1>0$.

1. Asymptotes: vertical asymptote $x=0$ remains. Oblique asymptote $y=-2x+3$ of $y=f(x)$ becomes asymptote $y=-2$ of $y=f'(x)$.

2. Map out the curve $y=f'(x)$ by tracing the gradient of $y=f(x)$ as $x$ increases. Note: $y=f(x)$ is strictly decreasing ⇒ $f'(x)<0$ for all $x$ in its domain ⇒ the curve $y=f'(x)$ lies below the $x$-axis.

Answer key

$a=-2$, $b=3$, $d=0$; native derivative sketch has both branches below $y=-2$.

Native diagrams