Question

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

Answer 1

There are no holes. The x-intercepts are `x =-2` and and `x=2/3`. The y-intercept is `4/9`. The vertical asymptote is `x=-3/2` and the horizontal asymptote is `y=-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:
  `(2+x)(2-3x)=0`

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