College Algebra, Chapter 4, 4.1, Section 4.1, Problem 6

The graph of a quadratic function $\displaystyle f(x) = \frac{-1}{2} x^2 - 2x + 6$ is given.

a.) Find the coordinates of the vertex


$
\begin{equation}
\begin{aligned}

f(x) =& \frac{-1}{2} x^2 - 2x + 6
&&
\\
\\
f(x) =& \frac{-1}{2} (x^2 + 4x) + 6
&& \text{Factor } \frac{-1}{2} \text{ from the $x$-terms}
\\
\\
f(x) =& \frac{-1}{2} (x^2 + 4x + 4) + 6 - \left( \frac{-1}{2} \right) (4)
&& \text{Complete the square: add 4 inside parentheses, subtract } \left( \frac{-1}{2} \right) (4) \text{ outside}
\\
\\
f(x) =& \frac{-1}{2} (x + 2)^2 + 8
&& \text{Standard form}

\end{aligned}
\end{equation}
$


Using the formula of standard form of a quadratic function

$f(x) = a(x - h)^2 + k$

We know that the vertex is at $(h,k)$. So the vertex of function $f$ is at $(-2, 8)$.

b.) Find the maximum or minimum value of $f$.

Since the parabola opens downward the maximum value of $f$ is $f(-2) = 8$.

c.) Find the domain and range of $f$.

Based from the graph, the domain of $f$ is $[-6, 2]$ and the range is $[0, 8]$.