# How do you find the end behavior of (5x^2-4x+4) / (3x^2+2x-4)?

Dec 4, 2016

#### Answer:

See explanation and graph.

#### Explanation:

y = (5x^2-4x+4)/((3(x-(-1+sqrt13)/3)(x-(-1-sqrt13)/3))

y-intercept ( x = 0 ) : $- 1$.

Vertical asymptotes: $\downarrow x = \left(\frac{- 1 \pm \sqrt{13}}{3}\right) \uparrow$

As $x \to \pm \infty , y \to \frac{5}{3}$

So, horizontal asymptote: $\leftarrow y = \frac{5}{3} \rightarrow$.

Interestingly, this asymptote cuts the graph in ${Q}_{1}$ at $x = \frac{16}{11}$.

Yet it is tangent at $x = \pm \infty$.

There are two turning points at x = 0.1309 ( in ${Q}_{4}$ ) and x = 2.1164

( in ${Q}_{1}$ ), wherein f' = 0.

There exists a point of inflexion for an x between 11/3 and 2.1164.

graph{y(3x^2+2x-4)-(5x^2-4x+4)=0 [-20, 20, -10, 10]}

Dec 7, 2016

#### Answer:

End behaviour describes what the graph is doing at the ends. It answers what the y values are doing as x values approach each of the ends.

#### Explanation:

Looking at the graph in the previous answer, we see there is a horizontal asymptote $y = \frac{5}{3}$ and two vertical asymptotes $x = \frac{- 1 + \sqrt{13}}{3}$ and $x = \frac{- 1 - \sqrt{13}}{3}$.

For end behavior, there are 6 ends to consider:
1) as $x \rightarrow \infty , y \rightarrow {\left(\frac{5}{3}\right)}^{-}$
2) as $x \rightarrow - \infty , y \rightarrow {\left(\frac{5}{3}\right)}^{+}$
3) as $x \rightarrow {\left(\frac{- 1 - \sqrt{13}}{3}\right)}^{-}$ , $y \rightarrow \infty$
4) as $x \rightarrow {\left(\frac{- 1 - \sqrt{13}}{3}\right)}^{+}$ , $y \rightarrow - \infty$
5) as $x \rightarrow {\left(\frac{- 1 + \sqrt{13}}{3}\right)}^{-}$, $y \rightarrow - \infty$
6) as $x \rightarrow {\left(\frac{- 1 + \sqrt{13}}{3}\right)}^{+}$, $y \rightarrow \infty$