Question

multiple choice 1 point
for the function $f(x)=\frac{x^{2}}{(x + 1)(x + 2)(x + 3)(x + 4)}$, where is (are) the vertical asymptotes located?
+1,+2,+3,+4
none of these
$\frac{x^{2}}{x + 1},\frac{x^{2}}{x + 2},\frac{x^{2}}{x + 3},\frac{x^{2}}{x + 4}$
0
-1,-2,-3,-4

Explanation:

Step1: Recall vertical asymptote condition

Vertical asymptotes occur where the denominator is \(0\) (and numerator is not \(0\)). Set \((x + 1)(x + 2)(x+3)(x + 4)=0\).

Step2: Solve the equation

Using the zero - product property: \(x+1 = 0\Rightarrow x=-1\); \(x + 2=0\Rightarrow x=-2\); \(x+3 = 0\Rightarrow x=-3\); \(x + 4=0\Rightarrow x=-4\). And for \(x=-1,-2,-3,-4\), the numerator \(x^{2}
eq0\).

Answer:

E. \(-1,-2,-3,-4\)