# How do you graph #y=(2x^2+x)/(x^2+1)# using asymptotes, intercepts, end behavior?

##### 1 Answer

#### Answer:

There is a horizontal asymptote at

#### Explanation:

Since we can't divide anything by zero, (How do you put anything into zero parts?) the function is undefined where the denominator equals zero. Therefore, we can set the denominator equal to zero and solve to find the vertical asymptotes:

This means that the graph will **never** touch the imaginary lines running through

Secondly, the horizontal asymptote: since the numerator and denominator have the same degree polynomial (the degree of a polynomial is the highest power of an exponent) we just divide the leading terms:

There is a horizontal asymptote at

Learn more about horizontal and slant asymptotes here .

Since you have two vertical asymptotes, your function will open the same way on both the right and left. Since the function is composed of a positive function divided by another positive function, and we know that a positive number divided by another positive number= a positive number, the function will open up on either side. The more specific end behavior is that as x approaches both

One more thing:

Hope that helps!!

Finally, here's the graph:

graph{(2x^2+x)/(x^2-1) [-9.9, 10.1, -4.06, 5.94]}