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

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

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Answer 1 · 2018-02-15T09:47:41

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

Horizontal asymptote: At $y = 0$ $y=0$

X-intercept: None

Y-intercept: At $y = 1$ $y=1$

Holes: None

Explanation:

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

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

$x^{2} + x - 2$ $x^2+x-2$

$x^{2} + 2 x - x - 2$ $x^2+2x-x-2$

$x (x + 2) - 1 (x + 2)$ $x(x+2)-1(x+2)$

$(x + 2) (x - 1)$ $(x+2)(x-1)$

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

So,

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

$x = 1, - 2$ $x=1,-2$

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

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

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

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

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

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

Therefore, there is no x-intercept.

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

Doing so,

$- \frac{2}{0^{2} + 0 - 2}$ $-2/(0^2+0-2)$

$- \frac{2}{- 2}$ $-2/-2$

$- (- 1)$ $-(-1)$

$1$ $1$

The y-intercept is at $y = 1$ $y=1$

There are no holes.

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

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Explanation:

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