# How do you sketch the graph #f(x)=2x^3-12x^2+18x-1#?

##### 1 Answer

**Step #1#: Determine the first derivative**

#f'(x) = 6x^2 - 24x + 18#

**Step #2#: Determine the critical numbers**

These will occur when the derivative equals

#0 = 6x^2 - 24x + 18#

#0 = 6(x^2 - 4x + 3)#

#0 = (x - 3)(x - 1)#

#x = 3 or 1#

**Step #3#: Determine the intervals of increase/decrease**

We select test points.

**Test point 1: #x = 0#**

Since this is positive, the function is uniformly increasing on

**Test point 2: #x = 2#**

Since this is negative, the function is decreasing on

I won't select a test point for

**Step #4#: Determine the second derivative **

This is the derivative of the first derivative.

#f''(x) = 12x - 24#

**Step #5#: Determine the points of inflection**

These will occur when

#0 = 12x- 24#

#0 = 12(x - 2)#

#x = 2#

**Step #6#: Determine the intervals of concavity**

Once again, we select test points.

**Test point #1#: #x = 1# **

#f''(1) = 12(1) - 24 = -12#

This means that

This also means that

**Step #7#: Determine the x/y- intercept**

The y-intercept is

#f(0) = 2(0)^3 - 12(0)^2 + 18x - 1 = -1#

If you can't connect the graph, you could always make a table of values. In the end, you should get a graph similar to the following. graph{2x^3- 12x^2 + 18x - 1 [-10, 10, -5, 5]}

Hopefully this helps!