# How do you find vertical, horizontal and oblique asymptotes for f(x) =( 10x^2)/(x-4)?

Jun 8, 2018

Vertical asymptote at $x = 4$, no horizontal asymptote and
slant asymptote is
$y = 10 x + 40$

#### Explanation:

$f \left(x\right) = \frac{10 {x}^{2}}{x - 4}$

The vertical asymptotes will occur at those values of $x$ for which

the denominator is equal to zero. $x - 4 = 0 \mathmr{and} x = 4$

Thus, the graph will have a vertical asymptote at $x = 4$

The degree of the numerator is $2$ and the degree of the

denominator is $1$. Since the larger degree occurs in the

numerator, the graph will have no horizontal asymptote.

Since the degree of the numerator is greater by margin of $1$

we have a slant asymptote , found by quotient on long division.

$f \left(x\right) = \frac{10 {x}^{2}}{x - 4} = \left(10 x + 40\right) + \frac{160}{x - 4}$

Slant asymptote is $y = 10 x + 40$. [Ans]