Answer
The only vertical asymptote is x=-2.
The only horizontal asymptote is y=2.
Work Step by Step
Before attempting to find asymptotes, first, we must make sure the function is in the lowest terms:
$R(x)=\dfrac{2x^2-50}{x^2-3x-10}$
$R(x)=\dfrac{2(x^2-25)}{(x-5)(x+2)}$
$R(x)=\dfrac{2(x-5)(x+5)}{(x-5)(x+2)}$
$R(x)=\dfrac{2x+10}{x+2} \text{ , } x\ne5$
To find vertical asymptotes, we must find the values that make the denominator equal zero. In this case:
$x+2=0$
$x=-2$
There are two cases to determine horizontal asymptotes. If the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote will be y = 0. If the degrees of both the numerator and denominator are the same, the horizontal asymptote is the ratio of the leading coefficients.
In this case, we can see that the degrees of both the numerator and denominator are the same, so the horizontal asymptote is the ratio $\frac{2}{1} \rightarrow y=2$
