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

Question

1611 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-13T12:28:08

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

#g(x)=(x^3-16 x)/(4 x^2-4 x)# 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]