Elementary Algebra

1. We set \(y=0\) and solve the equation

   \begin{align*} 4x-x^2 \amp =0 \amp \amp \blert{\text{Factor the left side.}}\\ x(4-x) \amp =0 \amp \amp \blert{\text{Set each factor equal to 0.}}\\ x=0~~~~4-x \amp = 0 \amp \amp \blert{\text{Solve each equation.}}\\ x = 0~~~~x \amp = 4 \end{align*}
   The \(x\)-intercepts are the points \((0,0)\) and \((4,0)\text{.}\)

2. We make a table of values that includes the \(x\)-intercepts.

   \(x\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)
   \(y\)	\(-5\)	\(0\)	\(3\)	\(4\)	\(3\)	\(0\)	\(-5\)

   We plot the points and connect them with a parabola, as shown in the figure.

   

In Example 1 we found that the \(x\)-intercepts of the graph are \(x=0\) and \(x=4\text{.}\) The \(x\)-coordinate of the vertex is the average of these two numbers.

The \(x\)-coordinate of the vertex is \(x=2\text{.}\) To find the \(y\)-coordinate of the vertex, we substitute \(x=\alert{2}\) into the equation of the parabola.

The coordinates of the vertex are \((2,4)\text{,}\) as you can see in the graph above.

First we find the \(x\)-intercepts: we substitute \(y=0\) into the equation, and solve for \(x\text{.}\)

The \(x\)-intercepts are the points \((7,0)\) and \((1,0)\text{.}\) Next, we locate the vertex of the graph. The \(x\)-coordinate of the vertex is the average of the \(x\)-intercepts, so

You can check that \(x=4\) is halfway between the two \(x\)-intercepts of the graph shown in the figure. To find the \(y\)-coordinate of the vertex, we evaluate the formula for the parabola at\(x=\alert{4}\text{.}\)

The vertex of the parabola is the point \((4,-9)\text{.}\) We can also find the \(y\)-intercept of the graph by substituting \(x=\alert{0}\) into the equation.

\begin{equation*} \alert{0}^2-8(\alert{0})+7=7 \end{equation*}

The \(y\)-intercept is the point \((0,7)\text{.}\) We plot the vertex and the intercepts and draw a smooth curve through them. The completed graph is shown at right.