SOLUTION: My question is if the vertex of f(x)=-x^2+bx+8 has y-coordinate 17 and is in the second quadrant, find b.

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Question 103792This question is from textbook Precalculus A GRAPHING APPROACH
: My question is if the vertex of f(x)=-x^2+bx+8 has y-coordinate 17 and is in the second quadrant, find b. This question is from textbook Precalculus A GRAPHING APPROACH

Found 2 solutions by jim_thompson5910, edjones:
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
Start with the given equation

To find the x-coordinate of the vertex, simply use this formula

So plug in a=-1

Simplify

So the x value of the vertex is

Now plug this value in, as well as f(x)=17, to solve for b

Plug in and f(x)=17

Square to get

Multiply

Multiply both sides by 4 to eliminate any fractions

Subtract 68 from both sides

Combine like terms

Factor the right side

Now set each factor equal to zero and solve

or

or

But since the vertex is in the second quadrant, the answer is

Notice if we plug in b=-6 into , we get

and if we graph the equation we get:

and we can see that the vertex has a y value at y=17 and it lies in the second quadrant. So our answer is verified.

Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website!
vertex=(h,k)=(h,17)
-(x-h)^2+17=-x^2+bx+8
We need a -9 to add to 17 to give 8. a has to be negative to give -x^2. h must be negative so that the vertex can be in the 2nd quadrant.
h=-3
-(x+3)^2=-(x^2+6x+9)+17
=-x^2-6x-9+17
=-x^2-6x+8
b=-6 Ans.
Ed