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

Feb 24, 2018

Look below

#### Explanation:

Okay, the best way to start this problem is listing the basic information of the equation (Basically, what you wrote in the problem):

x-intercept: This occurs when y=0. Remember that only the numerator can cause the equation to go to 0.

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

$x = 0$

The coordinate of the x-intercept is (0,0).

y-intercept: This occurs when x=0. Simple!

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

The coordinate of the y-intercept is (0,0).

Vertical Asymptotes: The function cannot exist if the denominator is equal to zero. Those values are where our vertical asymptotes are.

${x}^{2} + x - 2 = 0$

$\left(x + 2\right) \left(x - 1\right) = 0$

$x = - 2 , 1$

Horizontal Asymptotes: The function CAN cross the horizontal asymptote. A good way to think of the H.A is that it guides the end behavior of the function. Since the degree of the numerator (1) is less than the degree of the denominator (2), the horizontal asymptote is at $y = 0$.

Holes: These occur when you have the same factor in the numerator and the denominator. In this case, since you cannot factor an $x$ from the denominator, there is nothing to cancel out, so there are no holes.

Using this information, let's graph it!

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

Let's review our basic features in the graph. The graph's only intersection with the axis is at the (0,0), which matches our numbers. We can see that the graph "avoids" the lines $x = - 2$ and $x = 1$, our vertical asymptotes. The right and left ends of the graph come close to the line $y = 0$, our horizontal asymptote. Notice how the middle part of the graph is able to cross the horizontal asymptote. Finally, there are no holes.

Hope this helps!