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

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

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Answer 1 · 2018-02-15T09:47:41

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.

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

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Explanation:

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