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

2 Answers
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: #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]}

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 = 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#