Thread: Slant asymptote

Originally Posted by cjru

Hello guys! Can you give me an idea on how to find the slant asymptote of the fiven function
x^2 - 2x
F(x) = x +2

Thnaks in advance.

Divide the numerator by the denominator and drop the remainder. Just let y equal to the polynomial part of the quotient. See here.

Sorry my post is confusing. here is the right format

F(X) = X^2 - 2x / x +2

Can you check my work if I did it right. So when I divide the numerator by the denominator i get

X+2 / X^2 - 2x = x-4, R = 8

therefore my equation will be this : x-4 + 8/X^2 +1

Originally Posted by cjru

Can you check my work if I did it right. So when I divide the numerator by the denominator i get

X+2 / X^2 - 2x = x-4, R = 8

therefore my equation will be this : x-4 + 8/X^2 +1

The remainder is , but who cares?

The slant asymptote is

Thank you very much. But how does the graph look like?

Originally Posted by cjru

Thank you very much. But how does the graph look like?

If you don't have a graphing calculator you could just assign arbitrary values for x and solve for y.

Remember, this function also has a vertical asymptote at x = -2.

Or go here for an online graphing calculator.

Thank you very much. so which one is I am going to input in the calculator? Is it the original function which is x^2 - 2x/x+2 or this
x-4 + 8/X +2?

I solve for the x and y intercept and I got (0,0) & (2,0). are my intercepts correct?

Originally Posted by cjru

Thank you very much. so which one is I am going to input in the calculator? Is it the original function which is x^2 - 2x/x+2 or this
x-4 + 8/X +2?

I solve for the x and y intercept and I got (0,0) & (2,0). are my intercepts correct?

Yes, the upper branch has those intercepts.

You'll want to graph the original rational function. Depending on the graphing calculator you use, you may have to enter the equations for the asymptotes in order to see them.

Thank you very much again. i really appreciate your help.

God bless