Question 103792
{{{f(x)=-x^2+bx+8}}} Start with the given equation



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


{{{x=-b/(2a)}}}



{{{x=-b/(2(-1))}}} So plug in a=-1



{{{x=b/2}}} Simplify



So the x value of the vertex is {{{x=b/2}}}



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



{{{17=-(b/2)^2+b(b/2)+8}}} Plug in {{{x=b/2}}} and f(x)=17




{{{17=-b^2/4+b(b/2)+8}}} Square {{{b/2}}} to get {{{b^2/4}}}



{{{17=-b^2/4+b^2/2+8}}} Multiply



{{{68=-b^2+2b^2+32}}} Multiply both sides by 4 to eliminate any fractions



{{{0=-b^2+2b^2+32-68}}} Subtract 68 from both sides



{{{0=b^2-36}}} Combine like terms



{{{0=(b+6)(b-6)}}} Factor the right side



Now set each factor equal to zero and solve



{{{b+6=0}}} or {{{b-6=0}}}



{{{b=-6}}} or {{{b=6}}}



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



{{{b=-6}}}


Notice if we plug in b=-6 into {{{f(x)=-x^2+bx+8}}}, we get



{{{f(x)=-x^2-6x+8}}} and if we graph the equation we get:



{{{ graph( 500, 500, -6, 5, -2, 20, -x^2-6x+8) }}}


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.