Question

find the vertical asymptote(s) of the function, if possible.
$g(x) = \frac{3x - 2}{x^2 + 3x - 10}$
a $x = 3$
b $x = \frac{2}{3}$
c $x = -\frac{10}{3}$
d $x = -5$ and $2$
e no vertical asymptote

Explanation:

Step1: Factor the denominator

To find the vertical asymptotes of a rational function \( g(x)=\frac{3x - 2}{x^{2}+3x - 10} \), we first factor the denominator. We need two numbers that multiply to \(- 10\) and add to \(3\). The numbers are \(5\) and \(-2\). So, \(x^{2}+3x - 10=(x + 5)(x-2)\).

Step2: Find the values that make the denominator zero

Vertical asymptotes occur where the denominator is zero (and the numerator is not zero at those points). Set the factored denominator equal to zero: \((x + 5)(x - 2)=0\). Using the zero - product property, we have \(x+5 = 0\) or \(x - 2=0\). Solving \(x+5 = 0\) gives \(x=-5\), and solving \(x - 2=0\) gives \(x = 2\). Now, we check the numerator \(3x-2\) at \(x=-5\): \(3(-5)-2=-15 - 2=-17
eq0\). At \(x = 2\): \(3(2)-2=6 - 2 = 4
eq0\). So, the values \(x=-5\) and \(x = 2\) make the denominator zero and the numerator non - zero, so they are the vertical asymptotes.

Answer:

D. \(x=-5\) and \(2\)