Chapter 1.1, Problem 7GP

Solution

To find: The minimum value of a quadratic function.

The minimum value of the function $y\left(x\right)=4x^2+16x-3$ is -19.

Given data: The given function is $y=4x^2+16x-3$ .

Method/Formula used:

The vertex of the graph of the function $y\left(x\right)=a{x}^2+bx+c$ that represents a parabola is given by $h=-b/2a,k=y\left(-b/2a\right)$ . If $a>0$ , k is the minimum value of the function $y\left(x\right)$ .

Calculation:

The given function is $y\left(x\right)=4x^2+16x-3$ .

Compare it with $y\left(x\right)=a{x}^2+bx+c$ that gives $a=4,b=16,c=-3$ .

The x -coordinate of vertex of graph is

  $\begin{array}{c}h=\frac{-b}{2a}\\ =\frac{-\left(16\right)}{2\left(4\right)}\\ =-2\end{array}$

The y -coordinate of vertex of graph is

  $\begin{array}{c}k=y\left(-2\right)\\ y\left(2\right)=4{\left(-2\right)}^2+16\left(-2\right)-3\leftarrow \text{substitute}x=2\text{in}y\left(x\right)\\ =16-32-3\\ =-19\end{array}$

Thus, the minimum value of the function $y\left(x\right)=4x^2+16x-3$ is -19.