Chapter 1.2, Problem 35E

Solution

To find: The minimum or maximum value of a quadratic function.

The minimum value of the function $y=15{\left(x-25\right)}^2+130$ is 130 that occurs for $x=25$ .

Given data: The given quadratic function is $y=15{\left(x-25\right)}^2+130$ .

Method/Formula used:

The form of a quadratic function $y\left(x\right)=a{\left(x-h\right)}^2+k$ is known as vertex form that represents a parabola with vertex is $\left(h,k\right)$ . The maximum or minimum of the function $y\left(x\right)=a{\left(x-h\right)}^2+k$ ... (1)

If $a>0$ then k is minimum value and if $a<0$ then k is maximum value of the function $y\left(x\right)$ .

Calculation:

The given function is $y=15{\left(x-25\right)}^2+130$ . Comparing it with the equation (1), $a=15>0$ , $h=25$ , and $k=130$ . Therefore, the vertex of the graph of the given function is $\left(25,130\right)$ .

Since, $a>0$ , therefore $k=130$ is the minimum value of the function $y=15{\left(x-25\right)}^2+130$ , that occurs for $x=25$ .