How do you sketch the general shape of #f(x)=x^3-4x^2+7# using end behavior?

1 Answer
Jul 25, 2017

Answer:

Look below :)

Explanation:

Since the first power is odd the general shape of the graph is similar to that of #x^3#. So the end behavior is #x^3#. We can also find the y-intercept by replacing all the x-values with zeros:

#f(0)=(0)^3-4(0)^2+7#

#y=7#

All you really need to know to sketch the general shape is the know the end behavior and the y-intercept.

This is the graph of #x^3#
graph{x^3 [-10, 10, -5, 5]}

This is the graph of #y=x^3-4x^2+7#

As you can see the point #(0,7)# is the y-intercept.

graph{x^3-4x^2+7 [-11.29, 11.21, -3.11, 8.14]}