[tex]f(x)=-x^2+bx-14\\\\y_{max}=86[/tex] [tex]y_{max}\ \text{is in the vertex of the parabola}.[/tex] [tex]\text{The coordinates of the vertex of the parabola}\ f(x)=ax^2+bx+c:\\\\V\left(\dfrac{-b}{2a};\ \dfrac{-(b^2-4ac)}{4a}\right)[/tex] [tex]\text{We have:}\\\\f(x)=-x^2+bx-14\to a=-1;\ b=b;\ c=-14[/tex] [tex]\text{Substitute:}\\\\\dfrac{-\left(b^2-4\cdot(-1)\cdot(-14)\right)}{4\cdot(-1)}=86\\\\\dfrac{b^2-56}{4}=86\ \ \ \ |\cdot4\\\\b^2-56=344\ \ \ \ |+56\\\\b^2=400\to b=\pm\sqrt{400}\\\\b=-20\ \vee\ b=20[/tex] Answer: b = -20 or b = 20.
