Answer
$\lim _{x\rightarrow \infty }\dfrac {12x^{8}-3}{3x^{8}-2x^{7}}=4;\lim _{x\rightarrow -\infty }\dfrac {12x^{8}-3}{3x^{8}-2x^{7}}=4;y=4$
Work Step by Step
$\lim _{x\rightarrow \infty }\dfrac {12x^{8}-3}{3x^{8}-2x^{7}}=\dfrac {12-\dfrac {3}{x^{8}}}{3-\dfrac {2}{x}}=\dfrac {12-0}{3-0}=\dfrac {12}{3}=4$
$\lim _{x\rightarrow -\infty }\dfrac {12x^{8}-3}{3x^{8}-2x^{7}}=\dfrac {12-\dfrac {3}{x^{8}}}{3-\dfrac {2}{x}}=\dfrac {12-0}{3-0}=\dfrac {12}{3}=4$
So the horizontal asymptote will be $y=4$