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find a and b so that f(x)=x^2+ax+b has a local min at (6,-5) i found the derivative: f'=2x+a then i found a 0=2x+a 0=2(6)+a 0=1...Asked by kate
find a and b so that f(x)=x^2+ax+b has a local min at (6,-5)
i found the derivative:
f'=2x+a
then i found a
0=2x+a
0=2(6)+a
0=12+a
a=-12
but i can't seem to get the correct value for b. i plugged -5 into the original function, but my book says the answer is b=31. how do i get 31??
i found the derivative:
f'=2x+a
then i found a
0=2x+a
0=2(6)+a
0=12+a
a=-12
but i can't seem to get the correct value for b. i plugged -5 into the original function, but my book says the answer is b=31. how do i get 31??
Answers
Answered by
DanH
Here's how you get 31:
You are correct that a = -12.
Do, now the function becomes:
f(x) = x^2 - 12x + b.
Now, we know that x is 6 and y is -5. So plug these in:
-5 = 6^2 - 12(6) + b
solve for b and get 31.
Do you understand?
You are correct that a = -12.
Do, now the function becomes:
f(x) = x^2 - 12x + b.
Now, we know that x is 6 and y is -5. So plug these in:
-5 = 6^2 - 12(6) + b
solve for b and get 31.
Do you understand?
Answered by
kate
yes, i feel like such an idiot, but it makes sense now. thank you!!!
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