Question

find all vertical asymptotes of the following function.
$f(x) = \dfrac{2x^2 - 50}{x^2 + 2x}$

Explanation:

Step1: Find the denominator's roots

To find vertical asymptotes, we first set the denominator equal to zero: \(x^{2}+2x = 0\)
Factor the denominator: \(x(x + 2)=0\)

Step2: Solve for x

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we have two cases:
Case 1: \(x=0\)
Case 2: \(x + 2=0\), which gives \(x=-2\)

Step3: Check if numerator is non - zero at these x - values

For \(x = 0\): Substitute \(x = 0\) into the numerator \(2x^{2}-50\). We get \(2(0)^{2}-50=-50
eq0\)
For \(x=-2\): Substitute \(x=-2\) into the numerator \(2x^{2}-50\). We have \(2(-2)^{2}-50=2\times4 - 50=8 - 50=-42
eq0\)
Since the numerator is non - zero at \(x = 0\) and \(x=-2\) (the values that make the denominator zero), these are the vertical asymptotes.

Answer:

The vertical asymptotes are \(x = 0\) and \(x=-2\)