Question

what is the equation of the slant asymptote of the graph of the function $f(x) = \frac{2x^2 - 7x - 8}{x - 4}$? (1 point) \\(\circ\\) $y = 2x + 2$ \\(\circ\\) $y = x + 3$ \\(\circ\\) $y = 2x + 1$ \\(\circ\\) $y = 2x - 15$

Explanation:

Step1: Polynomial long division

Divide $2x^2 -7x -8$ by $x-4$:
First term: $\frac{2x^2}{x}=2x$. Multiply $x-4$ by $2x$: $2x^2-8x$.
Subtract: $(2x^2-7x-8)-(2x^2-8x)=x-8$.
Second term: $\frac{x}{x}=1$. Multiply $x-4$ by 1: $x-4$.
Subtract: $(x-8)-(x-4)=-4$.
Result: $f(x)=2x+1+\frac{-4}{x-4}$

Step2: Identify slant asymptote

As $x\to\pm\infty$, $\frac{-4}{x-4}\to0$. The slant asymptote is the polynomial part: $y=2x+1$

Answer:

y = 2x + 1