Question

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

$f\left(x\right)=3-x^2-6x$

Short answer

The function has maximum value that is 12.

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 $f\left(x\right)=3-x^2-6x=-x^2-6x+3$

Step 3. Solution.

In the function $f\left(x\right)=-x^2-6x+3$, we have $a=-1,b=-6,c=3$

Here, $a=-1<0$

So, the graph opens down and has a maximum value.

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

Thex-coordinate of the vertex is

$\begin{array}{l}\frac{-b}{2a}\\ =\frac{-\left(-6\right)}{2(-1)}\mathrm{..........}\left[a=-1,b=-6\right]\\ =-3\end{array}$

Find they-coordinate of the vertex by evaluating the function for $x=-3$.

$f(-3)=-{\left(-3\right)}^2-6\left(-3\right)+3=12$

So, the maximum value of function is 12.