How do you find the asymptotes for g(x)= (x^3 -16x )/ (4x^2 - 4x)g(x)=x316x4x24x?

1 Answer
Aug 13, 2018

Vertical asymptote: x=1x=1, slant asymptote : y=0.25x+0.25y=0.25x+0.25, removable discontinuity: x=0x=0

Explanation:

g(x)=(x^3-16 x)/(4 x^2-4 x)g(x)=x316x4x24x or

g(x)=(cancelx(x^2-16 ))/(4 cancelx(x-1))or

g(x)=(x^2-16)/(4x-4)

Vertical asymptote occur when denominator is zero.

x=0 is removable discontinuity. 4(x-1)=0

:.x=1 is vertical asymptote.

The numerator's degree is greater (by a margin of 1), then we

have a slant asymptote which is found doing long division.

(x^2-16)/(4x-4)= (0.25 x+0.25) -15/(4x-4)

Therefore slant asymptote is y= 0.25 x +0.25

graph{(x^3-16x)/(4x^2-4x) [-40, 40, -20, 20]}[Ans]