Sketch the graph of the rational function y = rac(x+2)(x−2)(x+1) using vertical and horizontal asymptotes, intercepts, and interval testing.

Vertical asymptotes occur where the denominator is zero and the numerator is non-zero.
Set the denominator to zero:

This gives $x-2=0$ or $x+1=0$.
Therefore, the vertical asymptotes are at $x=2$ and $x=-1$.

Compare the degree of the numerator ($n$) to the degree of the denominator ($m$).
The numerator is $x+2$, so its degree is $n=1$.
The denominator is $(x-2)(x+1) = x^2 - x - 2$, so its degree is $m=2$.
Since $n < m$, the horizontal asymptote is $y=0$.

X-intercepts occur where $y=0$. This happens when the numerator is zero (and the denominator is non-zero).
Set the numerator to zero:

The x-intercept is at $\boxed{(-2, 0)}$.

Y-intercepts occur where $x=0$. Substitute $x=0$ into the function:

The y-intercept is at $\boxed{(0, -1)}$.

Use the x-intercept ($x=-2$) and vertical asymptotes ($x=-1, x=2$) to divide the x-axis into intervals: $(-\infty, -2)$, $(-2, -1)$, $(-1, 2)$, $(2, \infty)$. Test a point in each interval to determine the sign of $y$.

• For $x < -2$ (e.g., $x=-3$): $y = \frac{-3+2}{(-3-2)(-3+1)} = \frac{-1}{(-5)(-2)} = \frac{-1}{10} < 0$. The graph is below the x-axis.
• For $-2 < x < -1$ (e.g., $x=-1.5$): $y = \frac{-1.5+2}{(-1.5-2)(-1.5+1)} = \frac{0.5}{(-3.5)(-0.5)} = \frac{0.5}{1.75} > 0$. The graph is above the x-axis.
• For $-1 < x < 2$ (e.g., $x=0$): $y = \frac{0+2}{(0-2)(0+1)} = \frac{2}{(-2)(1)} = -1 < 0$. The graph is below the x-axis.
• For $x > 2$ (e.g., $x=3$): $y = \frac{3+2}{(3-2)(3+1)} = \frac{5}{(1)(4)} = \frac{5}{4} > 0$. The graph is above the x-axis.

These points and behaviors allow you to sketch the curve, approaching the asymptotes and passing through the intercepts.