Question

find the x values where the rational function has a hole or vertical asymptote. 20. $y = \frac{(x + 3)(x - 2)}{(x + 3)^2(x - 2)}$ hole(s): blank vertical asymptote(s): blank

Explanation:

Step1: Simplify the rational function

Cancel out the common factors in the numerator and denominator. Given $y = \frac{(x + 3)(x - 2)}{(x + 3)^2(x - 2)}$, we can cancel out one factor of $(x + 3)$ and one factor of $(x - 2)$ (for $x
eq - 3$ and $x
eq2$). The simplified function is $y=\frac{1}{x + 3}$ for $x
eq2$.

Step2: Find the hole

A hole occurs when a factor cancels out in the numerator and denominator. Since $(x + 3)$ and $(x - 2)$ cancel out, we find the value of $x$ that makes the canceled - out factor zero. For the factor $(x + 3)$ and $(x - 2)$, setting $x+3 = 0$ gives $x=-3$ and setting $x - 2=0$ gives $x = 2$. But after simplification, the function is still undefined at $x = 2$. The hole occurs when the simplified function is well - defined at the value that made the original function undefined due to the canceled factor. So the hole is at $x=-3$.

Step3: Find the vertical asymptote

A vertical asymptote occurs at the value of $x$ that makes the denominator of the simplified non - reducible rational function equal to zero. For the function $y=\frac{1}{x + 3}$, setting the denominator $x+3=0$ gives $x=-3$. But since we already accounted for the hole at $x=-3$, and considering the original non - simplified form and the simplified form, the vertical asymptote is at $x = 2$ (because after canceling common factors, the remaining factor in the denominator $(x - 2)$ makes the function approach infinity as $x
ightarrow2$).

Answer:

Hole(s): $x=-3$
Vertical Asymptote(s): $x = 2$