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.