Question

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

Answer 1

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 `oo` when getting closer to `x=6` and on which side the `y` values approach `-oo` 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 + 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+4x-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 `(x^2+4x-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.