Answer
Vertical asymptotes: $x=2$ and $x=-\dfrac{7}{4}$
Horizontal asymptotes: $y=\dfrac{1}{2}$
Work Step by Step
$r(x)=\dfrac{(x+1)(2x-3)}{(x-2)(4x+7)}$
Vertical asymptotes
A rational function has vertical asymptotes where the function is undefined, that is, where the denominator is zero.
Set the denominator equal to 0 and solve for x to find the vertical asymptotes of this function:
$(x-2)(4x+7)=0$
$x-2=0$
$x=2$
$4x+7=0$
$4x=-7$
$x=-\dfrac{7}{4}$
Horizontal asymptote
Rewrite the function by evaluating the products in the numerator and in the denominator:
$r(x)=\dfrac{2x^{2}-3x+2x-3}{4x^{2}+7x-8x-14}=\dfrac{2x^{2}-x-3}{4x^{2}-x-14}$
The degrees of the numerator and the denominator are the same. Divide the leading coefficient of the numerator and the leading coefficient of the denominator to obtain the horizontal asymptote:
$y=\dfrac{2}{4}=\dfrac{1}{2}$