4.5: Graph Sketching and Asymptotes

Graph sketching depends on points of transition;
points where the sign of the slope changes position.


Problem:

1. Perform all necessary steps needed to sketch the graph of the function:
f(x)=x^3-3x^2

f'(x)=3x^2-6x =0
3x(x-2)=0
3x=0; x=0 & (x-2)=0;x=2

f(0)=0^3-3(0)^2 = 0
f(2)=2^3-3(2)^2 = -4

Critical Points: (0,0) & (2,-4)

For the sign graph of the first derivative: f'(-1)=9 - indicating a positive slope, f'(1)=-3 - indicating a negative slope and f'(3)=9 - indicating a positive slope.


f''(x)= 6x-6 =0
x=1
f(1)=1^3-3(1)^2 = -2

Point of Inflection: (1,-2)

For the sign graph of the second derivative: f''(0)=-6 - indicating a negative slope, f''(2)=6 - indicating a positive slope.

When the sign graph of the first and second derivatives are combined, the new sign graph looks like this: +-  0  --  1  -+  2  ++. The two zeros of the graph are located at 0 and 2 and the point of inflection, where the slope of the graph begins to change direction is at 1. Using the combined sign graphs from the first and second derivatives, I was able to draw a sketch of the graph of this function: