Question

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

Answer 1

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]}