Question

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

Answer 1

Below

Explanation:

`h(x) = (3x^2+2x-5)/(x-2)`

`h(x) = ((3x+5)(x-1))/(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(x) = ((x-2)(3x+8)+11)/(x-2)`

`h(x) = 3x+8+11/(x-2)`

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

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

`0=(3x^2+2x-5)/(x-2)`

`0=(3x+5)(x-1)`

`x=1, -5/3`
ie `(1,0)` `(-5/3,0)`

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

`y=(3x^2+2x-5)/(x-2)`

`y=(0+0-5)/(0-2)`

`y=-5/2`

ie `(0,-5/2)`

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