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
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