Question

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

Answer 1

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: `darr x =((-1+-sqrt13)/3) uarr`

As `x to +-oo, y to 5/3`

So, horizontal asymptote: `larr y = 5/3 rarr`.

Interestingly, this asymptote cuts the graph in `Q_1` at `x = 16/11`.

Yet it is tangent at `x = +-oo`.

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]}

Answer 2

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 = 5/3` and two vertical asymptotes `x=(-1+sqrt(13))/3` and `x = (-1-sqrt(13))/3`.

For end behavior, there are 6 ends to consider:
1) as `x rarr oo, y rarr (5/3)^-`
2) as `x rarr -oo, y rarr (5/3)^+`
3) as `x rarr ((-1-sqrt(13))/3)^-` , `y rarr oo`
4) as `x rarr ((-1-sqrt(13))/3)^+` , `y rarr -oo`
5) as `x rarr ((-1+sqrt(13))/3)^-`, `y rarr -oo`
6) as `x rarr ((-1+sqrt(13))/3)^+`, `y rarr oo`