How do you sketch the graph #y=x^3-4x^2# using the first and second derivatives?

Question

Emily Ammerman · Accepted Answer

# y=x^3-4x^2 # 1. General Observations A cubic function and the cubic coefficient is +ve, so it will have the classic cubic shape with a maximum to the left of a minimum,  2. Roots #y=0 => x^3 - 4x^2 = 0#
# :. x^2(x-4) = 0 #
# :. x=0 "(double root"), x=4 # So we can already deduce that there must be a local maximum at #x=0# that touches the axis (because of the double root) and a local minimum in between #x=0# and #x=4# (or else it could not be a cubic with a +ve cubic coefficient).  3. Turning Points # y=x^3-4x^2 => y' = 3x^2-8x #
At min/max #y'(0)=0 => 3x^2-8x = 0#
# :. x(3x-8) = 0#
# :. x=0 "(as expected)", x=8/3 (~~2.7)# When #x=0 => y=0#
When #x=8/3 => y= 512/27-256/9=-245/2 (~~-9.5)# 4. Nature of Turning Points # y' = 3x^2-8x => y''=6x-8# When #x=0 => y''<0 => "maximum (as expected)"#
When #x=8/3 => y''>0 => "minimum (as expected)"# 5. The Graph There is now information to plot the graph, here I will use the actual graph: graph{x^3-4x^2 [-10, 10, -15, 10]}

Elijah Abram · Answer

undefined To sketch the graph of $y = x^3 - 4x^2$ using the first and second derivatives:

1. Find the critical points by setting the first derivative equal to zero and solving for $x$.
2. Use the second derivative test to determine the nature of the critical points (whether they are local maxima, local minima, or points of inflection).
3. Identify any horizontal or vertical asymptotes, if applicable.
4. Plot the critical points and any asymptotes on the graph.
5. Determine the behavior of the function as $x$ approaches positive and negative infinity.
6. Sketch the graph based on the above information.

To summarize, you first find the critical points using the first derivative, then analyze the concavity using the second derivative, and finally incorporate additional information such as asymptotes and the behavior of the function at infinity to sketch the graph.