How do you graph y=(x^2+4x-5)/(x-6) using asymptotes, intercepts, end behavior?

Feb 11, 2018

Asymptotes:

A vertical asymptote occurs when the function is undefined so in this case, when $x = 6$.

We can figure out on which side the $y$ values approach $\infty$ when getting closer to $x = 6$ and on which side the $y$ values approach $- \infty$ by plugging in a slightly smaller number than 6 and a slightly bigger number than 6 for $x$.

If we plug in 6.00001 for $x$ we get a positive number for $y$ and if we plug in 5.9999 for $x$ we get a negative number for $y$.

So directly to the right of the asymptote the values are positive, and directly to the left they are negative.

In addition to the vertical asymptote at $x = 6$ there is a slant asymptote.

We know this because the magnitude of the numerator's function and the denominator's function differ by +1.

To find where the slant asymptote is, you need to divide the numerator by the denominator. This results in $x + 10 + \frac{55}{x - 6}$.

But the slant intercept doesn't include the remainder of the quotient. So there is a slant asymptote at $y = x + 10$

Intercepts

We calculate the intercepts of a rational function by finding the roots of the numerator's function.

So by factoring ${x}^{2} + 4 x - 5$ we will get the roots of the main function.

These roots are x = -5, 1.

End Behavior

When calculating end behavior you can rewrite the function with only it's leading terms. So our function $\frac{{x}^{2} + 4 x - 5}{x - 6}$ can be rewritten as ${x}^{2} / x$ which can be simplified to $x$.

Now, if we insert a large positive number we would get a large positive number, and similarly if we insert a large negative number we would get a large negative number.

So the end behavior is when the $x$ is a large positive number the $y$ will be a large positive number and when the $x$ is a large negative number the $y$ will be a large negative number.