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

Aug 13, 2018

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

Explanation:

$g \left(x\right) = \frac{{x}^{3} - 16 x}{4 {x}^{2} - 4 x}$ or

$g \left(x\right) = \frac{\cancel{x} \left({x}^{2} - 16\right)}{4 \cancel{x} \left(x - 1\right)}$or

$g \left(x\right) = \frac{{x}^{2} - 16}{4 x - 4}$

Vertical asymptote occur when denominator is zero.

$x = 0$ is removable discontinuity. $4 \left(x - 1\right) = 0$

$\therefore x = 1$ 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.

$\frac{{x}^{2} - 16}{4 x - 4} = \left(0.25 x + 0.25\right) - \frac{15}{4 x - 4}$

Therefore slant asymptote is $y = 0.25 x + 0.25$

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