What is the end behavior and turning points of #y = -2x^3 + 3x - 1#?

2 Answers
Sep 26, 2015

Answer:

As #x->oo, y->-oo and as x-> -oo, y->oo#

Turning points are #x= -1/sqrt2, +1/sqrt 2#.

Explanation:

For end behaviour, note the leading coefficient and the degree. The degree is odd and the leading coefficient negative, hence it would rise to the left and fall to the right.

For turning points get y'=0 and solve. In this case it is #-6x^2+3=0#
This gives #x= -1/sqrt2, +1/sqrt 2#.

Sep 26, 2015

To find the turning points, differentiate the function and set it to zero.
This yields :

#dy/dx=0iff -6x^2+3=0 iff x = +-1/4#

The corresponding y-values are then #y(-1/4)=-55/32# and #y(1/4)=-9/32#

To find the regions in which the function is increasing and decreasing, we investigate the sign of the first derivative around the turning points and to find the inflection point we set the second derivative to zero and find x = 0 and the corresponding y values is then #y(0)=-1#
To find the concavity, we investigate the sign of the second derivative on each side of the inflection point.

Since #y(1)=0=>x=1 # is a root and x intercept and so #(x-1) # is a factor. We may then long divide to find the other factors and roots.

Putting all together, we get the graph :

graph{-2x^3+3x-1 [-10, 10, -5, 5]}