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

1 Answer

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#