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

##### 1 Answer

Dec 9, 2016

There are no holes. The x-intercepts are

#### 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:

#(2+x)(2-3x)=0#

Then, using the zero product rule, set each factor equal to zero.

giving the x-intercepts

- The y-intercept can be found by setting x equal to 0 in the function.

So the y-intercept is

- The vertical asymptote(s) can be found by setting the denominator equal to zero:

- 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=(-3x^2-4x+4)/(4x^2+12x+9)#

Since the numerator and denominator both have degree 2, the equation of the horizontal asymptote is the ratio of the leading coefficients.

#y=(-3)/4#