Graph y=x^3-6x^2-96x

Problem

y=x3−6*x2−96*x

Solution

1. Find the y-intercept by setting x=0

y=0−6*(0)2−96*(0)=0

The graph passes through the origin (0,0)

2. Find the x-intercepts by setting y=0 and factoring the polynomial.

x*(x2−6*x−96)=0

Using the quadratic formula for the quadratic part:

x=(6±√(,36−4*(1)*(−96)))/2

x=(6±√(,420))/2=3±√(,105)

The x-intercepts are x=0 x≈13.25 and x≈−7.25

3. Find the first derivative to determine critical points and intervals of increase/decrease.

d(y)/d(x)=3*x2−12*x−96

4. Set the derivative to zero to find the x-coordinates of the local extrema.

3*(x2−4*x−32)=0

3*(x−8)*(x+4)=0

Critical points occur at x=8 and x=−4

5. Calculate the y-coordinates for the local extrema.
   For x=−4

y=(−4)3−6*(−4)2−96*(−4)=−64−96+384=224

For x=8

y=(8)3−6*(8)2−96*(8)=512−384−768=−640

Local maximum at (−4,224) and local minimum at (8,−640)

6. Find the second derivative to determine concavity and the inflection point.

d2(y)/(d(x)2)=6*x−12

7. Set the second derivative to zero to find the inflection point.

6*x−12=0⇒x=2

y=(2)3−6*(2)2−96*(2)=8−24−192=−208

The inflection point is at (2,−208)

8. Determine end behavior by looking at the leading term x3
   As x→∞ y→∞
   As x→−∞ y→−∞

Final Answer

y=x3−6*x2−96*x

The graph is a cubic curve with x-intercepts at x=0 and x=3±√(,105) a local maximum at (−4,224) a local minimum at (8,−640) and an inflection point at (2,−208)

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