College Algebra, Chapter 4, 4.1, Section 4.1, Problem 12

A quadratic function $f(x) = -x^2 + 10x$.

a.) Find the quadratic function in standard form.


$
\begin{equation}
\begin{aligned}

f(x) =& -x^2 + 10x
&&
\\
\\
f(x) =& -1(x^2 - 10x)
&& \text{Factor out $-1$ from each term}
\\
\\
f(x) =& -1 (x^2 - 10x + 25) - (1)(25)
&& \text{Complete the square: add 25 inside parentheses, subtract $(-1)(25)$ outside}
\\
\\
f(x) =& -(x - 5)^2 + 25
&& \text{Factor and simplify}

\end{aligned}
\end{equation}
$


The standard form is $f(x) = -(x - 5)^2 + 25$.

b.) Find its vertex and its $x$ and $y$-intercepts.

By using $f(x) = a (x - h)^2 + k$ with vertex at $(h,k)$.

The vertex of the function $f(x) = - (x - 5)^2 + 25$ is at $(5, 25)$.


$\begin{array}{llll}
\text{Solving for $x$-intercepts} & & \text{Solving for $y$-intercept} & \\
\text{We set } f(x) = 0, \text{ then} & & \text{We set } x = 0, \text{ then} & \\
0 = -(x - 5)^2 + 25 & \text{Add } (x - 5)^2 & y = -(0 - 5)^2 + 25 & \text{Substitute } x = 0 \\
(x - 5)^2 = 25 & \text{Take the square root} & y = - (-5)^2 + 25 & \text{Simplify} \\
x - 5 = \pm 5 & \text{Add } 5 & y = -25 + 25 & \text{Simplify} \\
x = \pm 5 + 5 & \text{Simplify} & y = 0 & \\
x = 0 \text{ and } x = 10 & & &
\end{array}
$


c.) Draw its graph.