Question 25057: Hello:The question is:
Find values of a, b, and c such that the graph of the polynomial p(x) = (ax^2) + bx + c passes through the point (-1,0) and has a horizontal tangent at (2, -9)
What I did first was substitute (-1, 0) into p(x): a4 + b2 + c = -9
Then I took the first derivative of p(x) and got: 2ax+b.
This is were I'm stuck. I believe I should use this for a second equation, then set up a matrix and solve for a,b and c, but I'm not sure.
Thanks
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! When you substitute (-1,0) into the form you should get the following:
a(-1)^2+b(-1)+c=0
a-b+c=0
When you substitute (2,-9) you should get:
4a+2b+c=9
The slope of your form is 2ax+b as you found when you
took the derivative.
That slope is horizontal, or zero, when x=2.
So you get 2a(2)+b=0
4a+b=0
You now have three equations involving a,b,and c.
Solve those as a system of equations in three variables
to find a,b, and c.
Cheers,
stan H.
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