Question

Determine whether each function has a maximum or minimum value. Then find the maximum or minimum value of the function.

$f\left(x\right)=4x^2+12x+9$

Short answer

The minimum value of function is 0.

Step-by-step solution

Step 1. Use the concept.

Consider the function $f\left(x\right)=a{x}^2+bx+c,a\ne 0$, the x-coordinate of vertex is $\frac{-b}{2a}$

The graph of $f\left(x\right)=a{x}^2+bx+c,a\ne 0$

• opens up and has a minimum value when $a>0$, and
• opens down and has a maximum value when $a<0$

Step 2. Given Information.

The given function is $f\left(x\right)=4x^2+12x+9$

Step 3. Solution.

In the function $f\left(x\right)=4x^2+12x+9$, we have

Here, $a=4>0$

So, the graph opens up and has a minimum value.

The minimum value of the function is the y-coordinate of the vertex.

Thex-coordinate of the vertex is

$\begin{array}{l}\frac{-b}{2a}\\ =\frac{-\left(12\right)}{2\left(4\right)}\mathrm{..........}\left[a=4,b=12\right]\\ =-1.5\end{array}$

Find they-coordinate of the vertex by evaluating the function for $x=-1.5$.

$f(-1.5)=4{\left(-1.5\right)}^2+12\left(-1.5\right)+9=0$

So, the minimum value of function is 0.