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??
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?
yes, i feel like such an idiot, but it makes sense now. thank you!!!
Well, let’s take a closer look at the problem. We know that the function has a local minimum at (6,-5).
If we substitute x = 6 into the original function, we get:
f(6) = (6^2) + a(6) + b = 36 + 6a + b
Since we know that the function has a local minimum at (6,-5), we can set f(6) = -5:
-5 = 36 + 6a + b
To find the value of b, we need to substitute the value of a that we found previously:
-5 = 36 + 6(-12) + b
-5 = 36 - 72 + b
-5 = -36 + b
Simplifying further, we get:
b = -5 - (-36)
b = 31
So, the correct value for b is indeed 31. Keep up the good work!
To find the correct value for b, you need to use the fact that the function has a local minimum at (6,-5). This means that when x = 6, the value of f(x) should be -5.
Using the original function f(x) = x^2 + ax + b, substitute x = 6 and f(x) = -5:
-5 = 6^2 - 12(6) + b
Simplifying, we get:
-5 = 36 - 72 + b
-5 = -36 + b
Add 36 to both sides:
31 = b
So, the correct value for b is 31.