This video is provided as supplementary

material for courses taught at Howard Community College and in this video I want to show how to find the equation

of a rational function when you’re given the asymptotes and the zeros. Here’s the first problem: Write an equation

for a rational function with vertical asymptotes at equals

-2 and 3, a horizontal asymptote at y=1, and zeros (1, 0) and (4, 0). So let’s call

this function f(x) and start out by using the vertical asymptotes to find

the factors of the denominator. We’ve got a vertical asymptote at x equals -2, which means that x + 2 will be a factor of the denominator. And I’ve

got a vertical asymptote at x=3, so x – 3 will be a factor of the denominator. Now I’ll

go on to find the factors of the numerator. I’ll get those by looking at the zeros. There’s a zero at (1, 0), so I’m going to

have x – 1 as a factor in the numerator. And there’s a zero at (4, 0) so I’ll have x – 4 as a factor in the numerator. Now let’s

take the binomials in the numerator and multiply those out.

And we’ll do the same for the denominator. So I’ll have f(x) equals x-squared and -5x and4. In the denominator I’ll have x-squared minus x minus 6. Now the last step is to check and make sure that

what I’ve got here will give me the right horizontal

asymptote. I want a horizontal asymptote aty=1. I’ve got two polynomials. They are both second-degree and they both have the same

leading coefficient, which would be one. So that means the

horizontal asymptote is going to be y=1, which is

what I want. So what I’ve got here is going to be the equation for the function that meets

these requirements. Let’s do one more. This says Write an equation for a rational

function with vertical asymptotes at x=4 and x=7, a horizontal

asymptote at y=-2, and zeros at (-3, 0) and (5, 0). So this is pretty much

the same process. I’ll call the function f(x) and we’ll find the

factors of the denominator by looking at the vertical asymptotes. I’ve got a vertical asymptote at x=4, so x – 4 will be a factor of the denominator. And I’ve got a vertical asymptote at x=7, so I also want x – 7 as a factor in the denominator. We’ll take a look at the zeros and find out

what the numerator will be. I’ve got a zero at (-3, 0 so I’ll have x + 3, and I’ve got (5, 0), so x – 5 will be a factor of the

numerator. We’ll multiply the binomials in the numerator and the in the denominator. That will give me f(x) equals x-squared – 2x – 15 as the

nominator. In the denominator, I’ll have x-squared – 11 x + 28. Now let’s check the horizontal asymptote. What I’ve got

here will have a horizontal asymptote y=1, since they’re both the second

degree polynomials with the same the coefficient. But I want y=-2 for my horizontal asymptote.

So I’ll take that numerator, the whole polynomial, and multiply it by -2. Let’s do that. I’m going to end up with f(x) equals -2x-squared + 4x + 30. The denominator is unchanged. That’s still x-squared – 11x + 28. So there’s my answer to that problem. And that’s really about all there is to it. Take care, I’ll see you next time.