How do you find the end behavior of 5x2−4x+43x2+2x−4?

2 Answers
A. S. Adikesavan
Dec 4, 2016

See explanation and graph.

Explanation:

y=5x2−4x+4(3(x−−1+√133)(x−−1−√133))

y-intercept ( x = 0 ) : −1.

Vertical asymptotes: ⏐⏐ ⏐ ⏐ ⏐↓x=(−1±√133)↑⏐ ⏐ ⏐ ⏐⏐

As x→±∞,y→53

So, horizontal asymptote: ←y=53→.

Interestingly, this asymptote cuts the graph in Q1 at x=1611.

Yet it is tangent at x=±∞.

There are two turning points at x = 0.1309 ( in Q4 ) and x = 2.1164

( in Q1 ), 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=53 and two vertical asymptotes x=−1+√133 and x=−1−√133.

For end behavior, there are 6 ends to consider:
1) as x→∞,y→(53)−
2) as x→−∞,y→(53)+
3) as x→(−1−√133)− , y→∞
4) as x→(−1−√133)+ , y→−∞
5) as x→(−1+√133)−, y→−∞
6) as x→(−1+√133)+, y→∞

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