Chapter 3, Problem 15RE

In Problems 15-17, determine whether the given quadratic function has a maximum value or a minimum value, and then find the value.

15. $f\left(x\right) = 3x^2 – 6x + 4$

To determine

To calculate: Whether the given function is having a maximum value or a minimum value and then finds the value.

Answer to Problem 15RE

The given has a minimum value which is at $y=1$.

Explanation of Solution

Given:

The given function is $f(x)=3x^2-6x+4$.

Formula used:

The general form of a quadratic equation is $f(x)=a{(x-h)}^2+k,a\ne 0$.

This general equation can also be written as $f(x)=a{x}^2+bx+c,a\ne 0$, where

$h=-\frac{b}{2a}$

and

$k=f\left(-\frac{b}{2a}\right)$

If $a>0$, the graph opens upwards.

If $a<0$, the graph opens downwards.

The vertex of the above function is $(h,k)=\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$.

If $a<0$, the vertex is the highest point on the graph and $f\left(-\frac{b}{2a}\right)$ is the maximum value of $f$.

If $a>0$, the vertex is the lowest point on the graph and $f\left(-\frac{b}{2a}\right)$ is the minimum value of $f$.

Calculation:

The given function is $f(x)=3x^2-6x+4$.

Here, we have $a=3$, $b=-6$ and $c=4$.

Therefore, we have $a>0$ and thus the vertex is the lowest point on the graph and the minimum value is

$f\left(-\frac{b}{2a}\right)=f\left(-\frac{-6}{2(3)}\right)=f(1)$

$=3{(1)}^2-6(1)+4$

$=3-6+4=1$