Question

sketch the graph of the rational function $f(x) = \frac{-(x + 2)^2}{x + 1}$

Explanation:

Step1: Find domain & vertical asymptote

The denominator cannot be zero: $x+1
eq0 \implies x
eq-1$. Vertical asymptote: $x=-1$.

Step2: Simplify the function

Use polynomial division or expand numerator:

$$\begin{align*} f(x)&=\frac{-(x^2+4x+4)}{x+1}\\ &=\frac{-(x+1)(x+3)-1}{x+1}\\ &=-(x+3)-\frac{1}{x+1}\\ &=-x-3-\frac{1}{x+1} \end{align*}$$

Step3: Find horizontal/slant asymptote

Since degree of numerator (2) is 1 more than denominator (1), there is a slant asymptote: $y=-x-3$.

Step4: Find x-intercepts

Set $f(x)=0$: $-(x+2)^2=0 \implies x=-2$. Intercept at $(-2,0)$.

Step5: Find y-intercept

Set $x=0$: $f(0)=\frac{-(0+2)^2}{0+1}=-4$. Intercept at $(0,-4)$.

Step6: Find holes

No common factors in numerator/denominator, so no holes.

Step7: Analyze sign & behavior

• For $x<-2$: $-(x+2)^2<0$, $x+1<0$, so $f(x)>0$
• For $-20$
• For $x>-1$: $-(x+2)^2<0$, $x+1>0$, so $f(x)<0$

As $x\to-1^-$, $f(x)\to+\infty$; as $x\to-1^+$, $f(x)\to-\infty$.
As $x\to\pm\infty$, $f(x)\to -x-3$.

Step8: Find critical points (optional)

Take derivative:

$$ f'(x)=\frac{-2(x+2)(x+1)+(x+2)^2}{(x+1)^2}=\frac{-(x+2)(x)}{(x+1)^2} $$

Set $f'(x)=0$: $x=0$ or $x=-2$.

• Local max at $x=-2$ (point $(-2,0)$)
• Local min at $x=0$ (point $(0,-4)$)

Answer:

To sketch the graph:

1. Draw vertical asymptote $x=-1$ (dashed line) and slant asymptote $y=-x-3$ (dashed line).
2. Plot intercepts $(-2,0)$ and $(0,-4)$.
3. For $x<-1$: The curve is positive, approaches $+\infty$ as $x\to-1^-$, approaches the slant asymptote as $x\to-\infty$, and touches the x-axis at $(-2,0)$.
4. For $x>-1$: The curve is negative, approaches $-\infty$ as $x\to-1^+$, approaches the slant asymptote as $x\to+\infty$, and has a local minimum at $(0,-4)$.