Question

vertical asymptotes of rational functions
where is the removable discontinuity of ( f(x) = \frac{x + 5}{x^2 + 3x - 10} ) located?
options: ( x = 5 ), ( x = -2 ), ( x = -5 ), ( x = 2 )

Explanation:

Step1: Factor the denominator

The denominator is \(x^{2}+3x - 10\). We factor it as \((x + 5)(x-2)\) since \(5\times(- 2)=-10\) and \(5+(-2)=3\). So the function becomes \(f(x)=\frac{x + 5}{(x + 5)(x-2)}\).

Step2: Identify removable discontinuity

A removable discontinuity (hole) occurs when a factor is common to both the numerator and the denominator. Here, the common factor is \(x + 5\). We set the common factor equal to zero: \(x+5 = 0\), which gives \(x=-5\). After canceling the common factor, the function is defined as \(f(x)=\frac{1}{x - 2}\) (for \(x
eq - 5\) and \(x
eq2\)), so the removable discontinuity is at \(x=-5\).

Answer:

\(x = - 5\)