# How do you graph y=(x^2-2x)/(2x+3) using asymptotes, intercepts, end behavior?

Aug 24, 2016

graph{(x^2-2x)/(2x+3) [-20, 20, -10, 10]}

#### Explanation:

$y = \frac{{x}^{2} - 2 x}{2 x + 3}$

a)
You can already tell that it has a vertical asymptote. Just let the denominator equal to zero and solve for $x$.
$2 x + 3 = 0$ so $x = - \frac{3}{2}$ (evident on the graph above).

b)
$y$ shows zero values when $x = 0$ and $x = 2$

c)
Next, if $x \to \pm \infty$, $y \to \pm \infty$
Roughly speaking;
$y = \frac{{\infty}^{2}}{\infty} = \infty$ and $y = {\infty}^{2} / - \infty = - \infty$
*Mathematicians hate seeing the infinity symbol used like this. I think it is because infinity cannot simply be seen as a number but more of a concept.

d)
Finally, find the turning points;
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{2 x + 3} ^ 2 \left({x}^{2} + 3 x - 3\right) = 0$
In other words; ${x}^{2} + 3 x - 3 = 0$
Using $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2}$ you will find $x = \frac{- 3 \pm \sqrt{21}}{2}$.

So far, you have found these important points:
-Asymptote $x = - \frac{3}{2}$
-Points $\left(0 , 0\right)$ and $\left(2 , 0\right)$
-$\left(x \to \infty , y \to \infty\right)$ and $\left(x \to - \infty , y \to - \infty\right)$
-Turning points at $x = \frac{- 3 \pm \sqrt{21}}{2}$