Chapter 3, Problem 25RE

To determine

Given quadratic function has a maximum value or minimum value.

Answer to Problem 25RE

Minimum value is $1$ .

Explanation of Solution

Given information:

  $f\left(x\right)=3x^2-6x+4$

Calculation:

Compare the $3x^2-6x+4$ with $f\left(x\right)=a{x}^2+bx+c$

So, $a=3,b=-6,c=4$

The graph of a quadratic function $f\left(x\right)=a{x}^2+bx+c$ opens up if $a>0$ and opens down if $a<0$ .

Function has $a=3$ which is greater than $0$ .

So graph is opens up.

In case $a<0$ vertex is a highest point on the graph. Also, in case $a>0$ , then the vertex is a lowest point on the graph.

Since $a>0$ , the vertex is a minimum point.

Minimum value is,

Minimum value of the function $f$ is $f\left(\frac{-b}{2a}\right)$ if the vertex is the lowest point.

Put $a=3,b=-6$ and then evaluate $\frac{-b}{2a}$ .

So minimum value is at,

  $\begin{array}{l}x=\frac{-b}{2a}\\ =\frac{-\left(-6\right)}{2\left(3\right)}\\ =1\end{array}$

Now evaluating function $\frac{-b}{2a}$ by putting $x=1$ in the function $f\left(x\right)=3x^2-6x+4$ to get minimum value.

  $\begin{array}{l}f\left(\frac{-b}{2a}\right)=f\left(1\right)\\ =3{\left(1\right)}^2-6\left(1\right)+4\\ =3-6+4\\ =1\end{array}$

Thus, minimum value is $1$ .