Chapter 1.1, Problem 5P

Solution

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

The function $y=-3x^2+9x+2$ has maximum value $8.75$ that occurs at $x=1.5$ .

Given data: The given function is $y=\frac{1}{2}{x}^2-3x+2$

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

  

Here coefficient of $x^2$ is $\frac{1}{2}>0$ , therefore, the function has minimum value at its vertex.

The y - coordinate of the vertex $\left(3,-2.5\right)$ is -5. Since, here coefficient of $x^2$ is $1/2>0$ , therefore, $8.75$ is the maximum value of the function $y=\frac{1}{2}{x}^2-3x+2$ that occurs at $x=3$ .