Answer
$$\lim _{x\rightarrow \infty }\dfrac {4x^{2}-7}{8x^{2}+5x+2}=\dfrac {1}{2};\lim _{x\rightarrow -\infty }\dfrac {4x^{2}-7}{8x^{2}+5x+2}=\dfrac {1}{2};y=\dfrac {1}{2}$$
Work Step by Step
$$\lim _{x\rightarrow \infty }\dfrac {4x^{2}-7}{8x^{2}+5x+2}=\dfrac {4-\dfrac {7}{x^{2}}}{8+\dfrac {5}{x}+\dfrac {2}{x^{2}}}=\dfrac {4-0}{8+0+0}=\dfrac {1}{2}$$
$$\lim _{x\rightarrow -\infty }\dfrac {4x^{2}-7}{8x^{2}+5x+2}=\dfrac {4-\dfrac {7}{x^{2}}}{8+\dfrac {5}{x}+\dfrac {2}{x^{2}}}=\dfrac {4-0}{8-0+0}=\dfrac {1}{2}$$
So the horizontal asymptote will be $y=\frac{1}{2}$
