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

##### 1 Answer
Apr 19, 2018

There is a horizontal asymptote at $y = 2$, vertical asymptotes at $x = \pm 1$, and $x$-intercepts at $x = 0 , - \frac{1}{2}$. There is also a $y$-intercept at $\left(0 , 0\right)$.

#### 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:
$0 = {x}^{2} - 1$
$1 = {x}^{2}$
$x = \pm 1$.
This means that the graph will never touch the imaginary lines running through $x = 1$ and $x = - 1$.

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: $\frac{2 {x}^{2}}{x} ^ 2 = 2$
There is a horizontal asymptote at $y = 2$.
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 $\pm \infty$, the function approaches 2.

One more thing: $x$-intercepts: Set the numerator eqaual to zero and solve to find the $x$-intercepts:
$2 {x}^{2} + x = 0$
$x \left(2 x + 1\right) = 0$
$x = 0 , - \frac{1}{2}$
Hope that helps!!

Finally, here's the graph:
graph{(2x^2+x)/(x^2-1) [-9.9, 10.1, -4.06, 5.94]}