How do you identify all asymptotes or holes for h(x)=(3x^2+2x-5)/(x-2)?

May 31, 2018

Below

Explanation:

$h \left(x\right) = \frac{3 {x}^{2} + 2 x - 5}{x - 2}$

$h \left(x\right) = \frac{\left(3 x + 5\right) \left(x - 1\right)}{x - 2}$

For vertical asymptote, $x - 2 = 0$ since your denominator cannot equal to 0.

$x - 2 = 0$
$x = 2$ is your vertical asymptote

Now we need to find the horizontal or oblique asymptote.

If you divide your polynomial, you will end up getting:

$h \left(x\right) = \frac{\left(x - 2\right) \left(3 x + 8\right) + 11}{x - 2}$

$h \left(x\right) = 3 x + 8 + \frac{11}{x - 2}$

Therefore, your oblique asymptote is $y = 3 x + 8$

To find your x-intercepts, let $y = 0$

$0 = \frac{3 {x}^{2} + 2 x - 5}{x - 2}$

$0 = \left(3 x + 5\right) \left(x - 1\right)$

$x = 1 , - \frac{5}{3}$
ie $\left(1 , 0\right)$ $\left(- \frac{5}{3} , 0\right)$

To find your y-intercepts, let $x = 0$

$y = \frac{3 {x}^{2} + 2 x - 5}{x - 2}$

$y = \frac{0 + 0 - 5}{0 - 2}$

$y = - \frac{5}{2}$

ie $\left(0 , - \frac{5}{2}\right)$

graph{((3x+5)(x-1))/(x-2) [-10, 10, -5, 5]}