How do you find the asymptotes for #f(x) = ((3x + 1)^2 * (4x - 3) )/ ((2x + 1) * (x-1)^2)#?

1 Answer

Answer:

vertical asymptotes: #x=-1/2,x=1#
horizontal asymptote: #f(x)=18#
oblique asymptote: does not exist

Explanation:

Finding the Vertical Asymptotes

Given the function,

#f(x)=((3x+1)^2*(4x-3))/((2x+1)*(x-1)^2)#

Cancel out any factors which appear in the numerator and denominator. Since there aren't any, set the denominator equal to #0# and solve for #x#. The values of #x# are your vertical asymptotes.

#(2x+1)*(x-1)^2=0#

#2x+1=0color(white)(XXXXXXXX)(x-1)^2=0#

#2x=-1color(white)(XXXXXXXXX)x-1=0#

#color(green)(|bar(ul(color(white)(a/a)color(black)(x=-1/2)color(white)(a/a)|)))color(white)(XXXXXX)color(green)(|bar(ul(color(white)(a/a)color(black)(x=1)color(white)(a/a)|)))#

Finding the Horizontal Asymptote

Given the function,

#f(x)=((3x+1)^2*(4x-3))/((2x+1)*(x-1)^2)#

Expand the brackets.

#f(x)=((9x^2+6x+1) * (4x-3))/((2x+1) * (x^2-2x+1))#

#f(x)=(36x^3+24x^2+4x-27x^2-18x-3)/(2x^3-4x^2+2x+x^2-2x+1)#

#f(x)=(color(darkorange)36x^3-3x^2-14x-3)/(color(purple)2x^3-3x^2+1)#

Since the degree of the numerator is the same as the degree of the denominator, divide the #color(darkorange)("leading coefficient")# of the leading term in the numerator by the #color(purple)("leading coefficient")# of the leading term in the denominator to determine the horizontal asymptote.

#f(x)=color(darkorange)36/color(purple)2#

#color(green)(|bar(ul(color(white)(a/a)color(black)(f(x)=18)color(white)(a/a)|)))#

Finding the Oblique Asymptote

Given,

#f(x)=((3x+1)^2*(4x-3))/((2x+1)*(x-1)^2)=(36x^color(red)3-3x^2-14x-3)/(2x^color(blue)3-3x^2+1)#

There would be a slant asymptote if the #color(red)("degree")# of the leading term in the #color(red)("numerator")# was #1# value larger than the #color(blue)("degree")# of the leading term in the #color(blue)("denominator")#. In your case, we see that the degree in both the numerator and denominator are equal.

#:.#, the slant asymptote does not exist.

graph{(36x^3+24x^2+4x-27x^2-18x-3)/(2x^3-4x^2+2x+x^2-2x+1) [-2, 2, -20, 20]}