Question

Determine without graphing whether the given quadratic function had a maximum value or a minimum value and then find the value. F(x)=3x^2+12x-4Is this equation minimum or maximum? Also what is the value?

Answer 1

$\begin{gathered} f(x)=ax^2+bx+c \\ \\ a>0\colon\text{parabola opens upward (has a minimum)} \\ a<0\colon\text{parabola opens downward (has a max}imum) \end{gathered}$

Given function:

$f(x)=3x^2+12x-4$

a = 3

As a is positive the parabola opens upward. It has a minimum value.

To find the minimum:

1. Find the value of x where the parabola has the minimum value, use the next formula:

$\begin{gathered} x=-\frac{b}{2a} \\ \\ x=-\frac{12}{2(3)}=-\frac{12}{6}=-2 \end{gathered}$

2. Use the value of x where the function has minimum value (step 1) to find the value of the function:

$\begin{gathered} f(-2)=3(-2)^2+12(-2)-4 \\ f(-2)=3(4)-24-4 \\ f(-2)=12-24-4 \\ f(-2)=-16 \end{gathered}$Then, the minimum value of the given function is -16 (f(-2)=-16)