The quadratic function f(x) = -2x^2 + kx - 18 has a maximum value of 14, where k is a posi… — GRE General Test

The quadratic function $f(x) = -2x^2 + kx - 18$ has a maximum value of $14$ , where $k$ is a positive constant. What is the value of $k$ ?

Answer: 16

The value of

k

16

The vertex of the parabola

f(x) = -2x^2 + kx - 18

is located at

x = \frac{k}{4}

. Evaluating

f\left(\frac{k}{4}\right)

gives the maximum value

\frac{k^2}{8} - 18

. Setting this expression equal to

14

leads to

\frac{k^2}{8} = 32

, so

k^2 = 256

. Taking the positive root as required by the problem statement yields

k = 16

Find the

x

-coordinate of the vertex of the quadratic function.

For

f(x) = -2x^2 + kx - 18

, we have

a = -2

b = k

, and

c = -18

. The vertex occurs at

x = -\frac{b}{2a} = -\frac{k}{2(-2)} = \frac{k}{4}

The maximum or minimum of any quadratic function

ax^2 + bx + c

occurs at its vertex, where

x = -\frac{b}{2a}

Evaluate the function at the vertex to determine the maximum value in terms of

k

f\left(\frac{k}{4}\right) = -2\left(\frac{k}{4}\right)^2 + k\left(\frac{k}{4}\right) - 18 = -2\left(\frac{k^2}{16}\right) + \frac{k^2}{4} - 18 = -\frac{k^2}{8} + \frac{k^2}{4} - 18 = \frac{k^2}{8} - 18

Substituting the vertex

x

-coordinate into

f(x)

yields the maximum value of the downward-opening parabola.

Set the maximum value expression equal to

14

and solve for

k^2

\frac{k^2}{8} - 18 = 14 \implies \frac{k^2}{8} = 32 \implies k^2 = 256.

The problem states that the maximum value of

f(x)

14

Solve for the positive constant

k

k = \sqrt{256} = 16.

Taking the square root of

256

gives

k = 16

k = -16

. Since

k

is given as a positive constant,

k = 16

Finding the extreme value of a quadratic function by locating its vertex

Question