Given $f ( x ) = \frac { x ^ { 2 } - 8 x + 16 } { x - 1 }$ , determine the equations of any slant and vertical asymptote.
A)slant: $y = x + 3$ ; vertical: $x = 0$
B)slant: $y = x - 1$ ; vertical: $x = - 3$
C)slant: $y = x + 3$ ; vertical: $x = - 3$
D)slant: $y = x - 7$ ; vertical: $x = 1$
E)slant: $y = x - 7$ ; vertical: none

Question

Quizplus · Accepted Answer

The vertical asymptote occurs where the denominator is zero, which is at $x = 1$. To find the slant asymptote, divide the numerator by the denominator. The quadratic $x^2 - 8x + 16$ simplifies to $(x-4)^2$, but for the slant asymptote, we focus on the division, which yields $x - 7$ as the linear term, hence the slant asymptote is $y = x - 7$.

Given f(x)=x2−8x+16x−1f ( x ) = \frac { x ^ { 2 } - 8 x + 16 } { x - 1 }f(x)=x−1x2−8x+16​

Given $f ( x ) = \frac { x ^ { 2 } - 8 x + 16 } { x - 1 }$