Question

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

$f\left(x\right)=\frac{3}{4}{x}^2-5x-2$

Short answer

The function has minimum value that is -10.333.

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 $f\left(x\right)=\frac{3}{4}{x}^2-5x-2$

Step 3. Solution.

In the function $f\left(x\right)=\frac{3}{4}{x}^2-5x-2$, we have $a=\frac{3}{4},b=-5,c=-2$

Here, $a=\frac{3}{4}>0$

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

$\begin{array}{l}\frac{-b}{2a}\\ =\frac{-\left(-5\right)}{2\left(\frac{3}{4}\right)}\mathrm{..........}\left[a=\frac{3}{4},b=-5\right]\\ =3.333\end{array}$

Find they-coordinate of the vertex by evaluating the function for $x=3.333$.

$f(3.333)=\frac{3}{4}{\left(3.333\right)}^2-5\left(3.333\right)-2=-10.333$

So, the minimum value of function is -10.333.