Find the values of b such that the function $f ( x ) = - x ^ { 2 } + b x - 16$ has the given maximum value 65. ?
A)±18
B)±22
C)±24
D)±20
E)±26

Question

Quizplus · Accepted Answer

The maximum value of a quadratic function $f(x) = ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$. For the given function $f(x) = -x^2 + bx - 16$, $a = -1$, $b = b$, and $c = -16$. The maximum value is given as 65. To find the value of $b$, we use the vertex form of a quadratic function, which gives the maximum value when $a < 0$. The vertex form is $f(x) = a(x - h)^2 + k$, where $k$ is the maximum value, and $h$ is the x-coordinate of the vertex. Since $k = 65$, we can set up the equation for the maximum value:$$65 = -\left(-\frac{b}{2(-1)}\right)^2 + b\left(-\frac{b}{2(-1)}\right) - 16$$Simplifying, we find:$$65 = -\frac{b^2}{4} + \frac{b^2}{2} - 16$$$$65 = \frac{b^2}{4} - 16$$$$65 + 16 = \frac{b^2}{4}$$$$81 = \frac{b^2}{4}$$$$324 = b^2$$$$b = \pm18$$Therefore, the correct choice is $b = \pm18$, which corresponds to choice A.

Find the Values of B Such That the Function f(x)=−x2+bx−16f ( x ) = - x ^ { 2 } + b x - 16f(x)=−x2+bx−16

Find the Values of B Such That the Function $f ( x ) = - x ^ { 2 } + b x - 16$