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

2 Answers
A. S. Adikesavan
Dec 4, 2016

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]}

Mrs. R
Dec 7, 2016

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

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