Given $f ( x ) = \frac { x ^ { 2 } - 4 x + 3 } { x - 7 }$ ,determine the equations of any slant and vertical asymptote.
A)slant: y = x - 3;vertical: x = 3
B)slant: y = x + 3;vertical: x = 7
C)slant: none;vertical: none
D)slant: y = x - 11;vertical: none
E)slant: y = x + 10;vertical: x = -4

Question

Quizplus · Accepted Answer

To find the slant asymptote, divide the numerator by the denominator. The division $x^2 - 4x + 3$ by $x - 7$ gives a quotient of $x + 3$ with a remainder, indicating a slant asymptote of $y = x + 3$. The vertical asymptote occurs where the denominator is zero, which is at $x = 7$, hence the vertical asymptote is $x = 7$.

Given f(x)=x2−4x+3x−7f ( x ) = \frac { x ^ { 2 } - 4 x + 3 } { x - 7 }f(x)=x−7x2−4x+3​

Given $f ( x ) = \frac { x ^ { 2 } - 4 x + 3 } { x - 7 }$