Chapter 1.1, Problem 6P

Solution

To find: The minimum or maximum value of a quadratic function and determine the value of coordinate where it occurs.

The function $y=-\frac{3}{8}{x}^2+6x-5$ has maximum value $19$ that occurs at $x=8$ .

Given data: The given function is $y=-\frac{3}{8}{x}^2+6x-5$ .

Method/Formula used:

The graph of the function $y\left(x\right)=a{x}^2+bx+c$ represents a parabola whose y -coordinate of the vertex $\left(h,k\right)$ say is the minimum or maximum value of the function $y\left(x\right)$ and if $a<0$ , k is the maximum value and if $a>0$ then k is the minimum value of the function $y\left(x\right)$ . The vertex can be determined using graphing utility.

Calculation:

The given function is $y=-\frac{3}{8}{x}^2+6x-5$ . The vertex of the graph of the function $y\left(x\right)$ is shown in Fig. (1).

  

Here coefficient of $x^2$ is $-\frac{3}{8}<0$ , therefore, the function has maximum value at its vertex.

The y - coordinate of the vertex $\left(8,19\right)$ is -5. Since, here coefficient of $x^2$ is $-\frac{3}{8}<0$ , therefore, $19$ is the maximum value of the function $y=-\frac{3}{8}{x}^2+6x-5$ that occurs at $x=8$ .