SOLUTION: The graph of y = 2x^3 + ax^2 + b has a stationary point (-3,19) . Find the value of a and b. Determine the nature of the stationary point (-3,19).

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Question 1192428: The graph of y = 2x^3 + ax^2 + b has a stationary point (-3,19) . Find the value of a and b. Determine the nature of the stationary point (-3,19).
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


f%28x%29=2x%5E3%2Bax%5E2%2Bb

f%28-3%29=19
-54%2B9a%2Bb=19

df%2Fdx=6x%5E2%2B2ax

The stationary point at x=-3 is where the derivative is zero.

6x%5E2%2B2ax=54-6a=0
a=9

Use that value of a to solve for b.

-54%2B9%289%29%2Bb=19
27%2Bb=19
b=-8

The function is 2x%5E3%2B9x%5E2-8

We know f(-3)=19; and f(0)=-8. Those two points, along with the positive leading coefficient, tell us that the stationary point at (-3,19) is a local maximum.

Or, more formally, to show that the stationary point is a local maximum, we could find the second derivative of the function and show that it is negative at x=-3.

ANSWERS:
a=9
b=-8
(-3,19) is a local maximum

A graph....

graph%28400%2C400%2C-5%2C3%2C-10%2C30%2C2x%5E3%2B9x%5E2-8%29