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

Apr 30, 2016

vertical asymptotes: $x = - \frac{1}{2} , x = 1$
horizontal asymptote: $f \left(x\right) = 18$
oblique asymptote: does not exist

#### Explanation:

Finding the Vertical Asymptotes

Given the function,

$f \left(x\right) = \frac{{\left(3 x + 1\right)}^{2} \cdot \left(4 x - 3\right)}{\left(2 x + 1\right) \cdot {\left(x - 1\right)}^{2}}$

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.

$\left(2 x + 1\right) \cdot {\left(x - 1\right)}^{2} = 0$

$2 x + 1 = 0 \textcolor{w h i t e}{X X X X X X X X} {\left(x - 1\right)}^{2} = 0$

$2 x = - 1 \textcolor{w h i t e}{X X X X X X X X X} x - 1 = 0$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x = - \frac{1}{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \textcolor{w h i t e}{X X X X X X} \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x = 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Finding the Horizontal Asymptote

Given the function,

$f \left(x\right) = \frac{{\left(3 x + 1\right)}^{2} \cdot \left(4 x - 3\right)}{\left(2 x + 1\right) \cdot {\left(x - 1\right)}^{2}}$

Expand the brackets.

$f \left(x\right) = \frac{\left(9 {x}^{2} + 6 x + 1\right) \cdot \left(4 x - 3\right)}{\left(2 x + 1\right) \cdot \left({x}^{2} - 2 x + 1\right)}$

$f \left(x\right) = \frac{36 {x}^{3} + 24 {x}^{2} + 4 x - 27 {x}^{2} - 18 x - 3}{2 {x}^{3} - 4 {x}^{2} + 2 x + {x}^{2} - 2 x + 1}$

$f \left(x\right) = \frac{\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{36} {x}^{3} - 3 {x}^{2} - 14 x - 3}{\textcolor{p u r p \le}{2} {x}^{3} - 3 {x}^{2} + 1}$

Since the degree of the numerator is the same as the degree of the denominator, divide the $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{leading coefficient}}$ of the leading term in the numerator by the $\textcolor{p u r p \le}{\text{leading coefficient}}$ of the leading term in the denominator to determine the horizontal asymptote.

$f \left(x\right) = \frac{\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{36}}{\textcolor{p u r p \le}{2}}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{f \left(x\right) = 18} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Finding the Oblique Asymptote

Given,

$f \left(x\right) = \frac{{\left(3 x + 1\right)}^{2} \cdot \left(4 x - 3\right)}{\left(2 x + 1\right) \cdot {\left(x - 1\right)}^{2}} = \frac{36 {x}^{\textcolor{red}{3}} - 3 {x}^{2} - 14 x - 3}{2 {x}^{\textcolor{b l u e}{3}} - 3 {x}^{2} + 1}$

There would be a slant asymptote if the $\textcolor{red}{\text{degree}}$ of the leading term in the $\textcolor{red}{\text{numerator}}$ was $1$ value larger than the $\textcolor{b l u e}{\text{degree}}$ of the leading term in the $\textcolor{b l u e}{\text{denominator}}$. In your case, we see that the degree in both the numerator and denominator are equal.

$\therefore$, 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]}