# How do you identify all asymptotes or holes and intercepts for f(x)=((2+x)(2-3x))/(2x+3)^2?

Dec 9, 2016

There are no holes. The x-intercepts are $x = - 2$ and and $x = \frac{2}{3}$. The y-intercept is $\frac{4}{9}$. The vertical asymptote is $x = - \frac{3}{2}$ and the horizontal asymptote is $y = - \frac{3}{4}$

#### Explanation:

graph{y(2x+3)^2-(2+x)(2-3x)=9 [-10, 10, -5, 5]} - There are no holes since there are no factors that cancel out of both the numerator and denominator.

• The x-intercepts can be found by setting the numerator equal to zero:
$\left(2 + x\right) \left(2 - 3 x\right) = 0$

Then, using the zero product rule, set each factor equal to zero.
$2 + x = 0$ and $2 - 3 x = 0$
giving the x-intercepts
$x = - 2$ and $x = \frac{2}{3}$
- The y-intercept can be found by setting x equal to 0 in the function.
$f \left(0\right) = \frac{\left(2\right) \left(2\right)}{{3}^{2}}$
$f \left(0\right) = \frac{4}{9}$
So the y-intercept is $\frac{4}{9}$
- The vertical asymptote(s) can be found by setting the denominator equal to zero:
${\left(2 x + 3\right)}^{2} = 0$
$2 x + 3 = 0$
$2 x = - 3$
$x = - \frac{3}{2}$

• The horizontal asymptote can be found by looking at the degree and leading coefficients of the numerator and denominator.
Expanding the brackets in the function:
$y = \frac{- 3 {x}^{2} - 4 x + 4}{4 {x}^{2} + 12 x + 9}$
Since the numerator and denominator both have degree 2, the equation of the horizontal asymptote is the ratio of the leading coefficients.
$y = \frac{- 3}{4}$