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

1 Answer

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