Chapter 2.2, Problem 39E

To determine

To find: The minimum and maximum value of the function with the description of domain and range of the function.

Answer to Problem 39E

The function $y=6x^2-1$ opens up and has the minimum value $-1$ .

Domain is all the real numbers and ranges is greater than or equal to $-1,y\ge -1$ .

The given function $y=6x^2-1$ is decreasing to the left of $x=0$ and increasing to the right of $x=0$ .

Explanation of Solution

Given information: $y=6x^2-1$ .

General form of quadratic function is $y=a{x}^2+bx+c$ .

  $a=6,b=0,c=-1$

The coefficient of $x^2$ , $a>0$ so the parabola opens up and has a minimum value,

The coordinates of the vertex of the function gives the minimum value,

  $\begin{array}{l}x=-\frac{b}{2a}\\ x=-\frac{0}{2\times 6}\\ x=0\end{array}$

Substitute $x=0\text{in}y=6x^2-1$ ,

  $\begin{array}{l}$ y=6{(0)}^2-1$y=0-1\\ y=-1\end{array}$

The minimum value of the graph of the function $y=6x^2-1$ is $-1$ .

The domain $(x)$ is all the real numbers and the range $(y)$ is greater than or equal to $-1$ .

The given function $y=6x^2-1$ is decreasing to the left of $x=0$ and increasing to the right of $x=0$ .