An arrow is shot straight up into the air from 5 feet above the ground with a velocity of 18 ft/s. The quadratic expression that represents this situation is \begin{align*}5 + 18t  16t^2\end{align*}
Guidance
A graphing calculator can be a very helpful tool when graphing parabolas. This concept outlines how to use the TI83/84 to graph and find certain points on a parabola.
Example A
Graph \begin{align*}y=3x^2+14x8\end{align*}
Solution: Using a TI83/84, press the \begin{align*}Y=\end{align*}
If your graph does not look like this one, there may be an issue with your window. Press ZOOM and then 6:ZStandard, ENTER. This should give you the standard window.
Example B
Using your graphing calculator, find the vertex of the parabola from Example A.
Solution: To find the vertex, press \begin{align*}2^{nd}\end{align*}
Example C
Using your graphing calculator, find the \begin{align*}x\end{align*}
Solution: To find the \begin{align*}x\end{align*}
NOTE: When graphing parabolas and the vertex does not show up on the screen, you will need to zoom out. The calculator will not find the value(s) of any \begin{align*}x\end{align*}
Intro Problem Revisit Use your calculator to find the vertex of the parabolic expression \begin{align*}5 + 18t  16t^2\end{align*}
The vertex is (0.5625, 10.0625). Therefore, the maximum height is reached at 0.5625 seconds and that maximum height is 10.0625 feet.
Guided Practice
1. Graph \begin{align*}y=6x^2+11x35\end{align*}
Answers
1. Using the steps above, the vertex is (0.917, 40.04) and is \begin{align*}a\end{align*}
Explore More
Graph the quadratic equations using a graphing calculator. Find the vertex and \begin{align*}x\end{align*}

\begin{align*}y=x^2x6\end{align*}
y=x2−x−6 
\begin{align*}y=x^2+3x+28\end{align*}
y=−x2+3x+28 
\begin{align*}y=2x^2+11x40\end{align*}
y=2x2+11x−40 
\begin{align*}y=x^26x+7\end{align*}
y=x2−6x+7 
\begin{align*}y=x^2+8x+13\end{align*}
y=x2+8x+13 
\begin{align*}y=x^2+6x+34\end{align*}
y=x2+6x+34 
\begin{align*}y=10x^213x3\end{align*}
y=10x2−13x−3 
\begin{align*}y=4x^2+12x3\end{align*}
y=−4x2+12x−3 
\begin{align*}y=\frac{1}{3}(x4)^2+12\end{align*}
y=13(x−4)2+12 
\begin{align*}y=2(x+1)^29\end{align*}
y=−2(x+1)2−9
Calculator Investigation The parent graph of a quadratic equation is \begin{align*}y=x^2\end{align*}
 Graph \begin{align*}y=x^2, y=3x^2\end{align*}
y=x2,y=3x2 , and \begin{align*}y=\frac{1}{2}x^2\end{align*} on the same set of axes in the calculator. Describe how \begin{align*}a\end{align*} effects the shape of the parabola.  Graph \begin{align*}y=x^2, y=x^2\end{align*}, and \begin{align*}y=2x^2\end{align*} on the same set of axes in the calculator. Describe how \begin{align*}a\end{align*} effects the shape of the parabola.
 Graph \begin{align*}y=x^2, y=(x1)^2\end{align*}, and \begin{align*}y=(x+4)^2\end{align*} on the same set of axes in the calculator. Describe how \begin{align*}h\end{align*} effects the location of the parabola.
 Graph \begin{align*}y=x^2, y=x^2+2\end{align*}, and \begin{align*}y=x^25\end{align*} on the same set of axes in the calculator. Describe how \begin{align*}k\end{align*} effects the location of the parabola.
 Real World Application The path of a baseball hit by a bat follows a parabola. A batter hits a home run into the stands that can be modeled by the equation \begin{align*}y=0.003x^2+1.3x+4\end{align*}, where \begin{align*}x\end{align*} is the horizontal distance and \begin{align*}y\end{align*} is the height (in feet) of the ball. Find the maximum height of the ball and its total distance travelled.