Question

Describe the vertical asymptotes and holes for the graph of y=x-4/x^2+3x+2

Answer 1

Step 1: Factor the denominator

Answer 2

(x+2)(x+1)

Answer 3

vertical asymptotes for a rational expression occur when the fraction becomes undefined, and that only happens when the denominator turns to 0

thus \(\bf y=\cfrac{x-4}{{\color{brown}{ x^2+3x+2}}}\qquad x^2+3x+2=0\implies y=\cfrac{x-4}{0}
\\ \quad \\
\textit{thus the vertical asymptotes occur at }{\color{brown}{ x^2+3x+2}}=0\)

and you're correct, are (x+2)(x+1)

Answer 4

which means \(\large {
x^2+3x+2=0\\ \quad \\\implies (x+2)(x+1)=0\to
\begin{cases}
x+2=0\to x=-2\\
x+1=0\to x=-1
\end{cases}
\\ \quad \\
asymptotes\to
\begin{cases}
y=-2\\
y=-1
\end{cases}
}\)

Answer 5

and the "holes" occur if x = -2 or x = -1

Answer 6

hmmmm wait a second... is a vertical asymptote... so the equation should be x-based... so \(\large {x^2+3x+2=0\\ \quad \\ (x+2)(x+1)=0\to
\begin{cases}
x+2=0\to x=-2\\
x+1=0\to x=-1
\end{cases}
\\ \quad \\
asymptotes\ and \ holes\to
\begin{cases}
x=-2\\
x=-1
\end{cases}}\)

Answer 7

So there are holes because x=-2 and x=-1?

Answer 8

yes, because if we set x = -2 or -1
our denominator will turn to 0 and thus our equation will be undefined \(y=\cfrac{x-4}{{\color{brown}{ x^2+3x+2}}}\qquad x^2+3x+2=0\implies y=\cfrac{x-4}{0}
\\ \quad \\
y=\cfrac{x-4}{{\color{brown}{ (x+2)(x+1)}}}
\begin{cases}
y=\cfrac{x-4}{{\color{brown}{ ({\color{blue}{ -2}}+2)(x+1)}}}\to y=\cfrac{x-4}{0(x+1)}\to y=\cfrac{x-4}{0}
\\ \quad \\
y=\cfrac{x-4}{{\color{brown}{ (x+2)({\color{blue}{ -1}}+1)}}}\to y=\cfrac{x-4}{(x+2)0}\to y=\cfrac{x-4}{0}
\end{cases} \)