Question

find the x values where the rational function has a hole or vertical asymptote.

1. $r(x)=\frac{x^{3}-x^{2}}{x^{2}+2x + 1}$

hole(s)
vertical asymptote(s)

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{3}-x^{2}=x^{2}(x - 1)$. The denominator $x^{2}+2x + 1=(x + 1)^{2}$ since $a^{2}+2ab+b^{2}=(a + b)^{2}$ with $a=x$ and $b = 1$. There are no common - factors, so there are no holes.

Step2: Find the vertical asymptotes

Set the denominator equal to zero. $(x + 1)^{2}=0$. Solving for $x$, we get $x=-1$.

Answer:

Hole(s): None
Vertical Asymptote(s): $x=-1$