Question

How do you graph `f(x)=-2/(x^2+x-2)` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

Vertical asymptote: At `x=1,-2`

Horizontal asymptote: At `y=0`

X-intercept: None

Y-intercept: At `y=1`

Holes: None

Explanation:

To find the holes and vertical asymptotes, we first study the denominator.

We have `x^2+x-2` in the denominator. Factorizing this will give us the vertical asymptotes.

`x^2+x-2`

`x^2+2x-x-2`

`x(x+2)-1(x+2)`

`(x+2)(x-1)`

The asymptotes are found when the answer is `0`.

So,

`(x+2)(x-1)=0`

`x=1,-2`

There are vertical aymptotes at `x=1,-2`.

To find the horizontal asymptotes, we must look at the degree of the numerator (`n`) and the denominator (`m`).

If `n>m,` there is no horizontal asymptote

If `n=m`, we divide the leading coefficients,

If `n<m`, the asymptote is at `y=0`.

Here, `n=0` and `m=2`, so the horizontal asymptote is at `y=0`.

Therefore, there is no x-intercept.

The y-intercept is found by taking all `x` to be equal to `0`.

Doing so,

`-2/(0^2+0-2)`

`-2/-2`

`-(-1)`

`1`

The y-intercept is at `y=1`

There are no holes.