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*}, where *t* is the time in seconds. At what time does the arrow reach its maximum height and what is that height?

### Guidance

A graphing calculator can be a very helpful tool when graphing parabolas. This concept outlines how to use the TI-83/84 to graph and find certain points on a parabola.

#### Example A

Graph \begin{align*}y=-3x^2+14x-8\end{align*} using a graphing calculator.

**Solution:** Using a TI-83/84, press the \begin{align*}Y=\end{align*} button. Enter in the equation. Be careful not to confuse the negative sign and the subtraction sign. The equation should look like \begin{align*}y=-3x^2+14x-8\end{align*} or \begin{align*}y=-3x^2+14x-8\end{align*}. Press GRAPH.

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*} TRACE (CALC). The Calculate menu will appear. In this case, the vertex is a maximum, so select **4:maximum,** ENTER. The screen will return to your graph. Now, you need to tell the calculator the Left Bound. Using the arrows, arrow over to the left side of the vertex, press ENTER. Repeat this for the Right Bound. The calculator then takes a guess, press ENTER again. It should give you that the maximum is \begin{align*}X=2.3333333\end{align*} and \begin{align*}Y=8.3333333\end{align*}. As fractions, the coordinates of the vertex are \begin{align*}\left(2\frac{1}{3}, 8\frac{1}{3}\right)\end{align*}. Make sure to write the coordinates of the vertex as a point.

#### Example C

Using your graphing calculator, find the \begin{align*}x-\end{align*}intercepts of the parabola from Example A.

**Solution:** To find the \begin{align*}x-\end{align*}intercepts, press \begin{align*}2^{nd}\end{align*} TRACE (CALC). The Calculate menu will appear. Select **2:Zero,** ENTER. The screen will return to your graph. Let’s focus on the left-most intercept. Now, you need to tell the calculator the Left Bound. Using the arrows, arrow over to the left side of the vertex, press ENTER. Repeat this for the Right Bound (keep the bounds close to the intercept). The calculator then takes a guess, press ENTER again. This intercept is \begin{align*}X=.666667\end{align*}, or \begin{align*}\left(\frac{2}{3}, 0\right)\end{align*}. Repeat this process for the second intercept. You should get (4, 0).

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*}intercepts or the vertex that do not appear on screen. To zoom out, press ZOOM, **3:Zoom Out**, ENTER, ENTER.

**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+11x-35\end{align*} using a graphing calculator. Find the vertex and \begin{align*}x-\end{align*}intercepts. Round your answers to the nearest hundredth.

#### Answers

1. Using the steps above, the vertex is (-0.917, -40.04) and is \begin{align*}a\end{align*} *minimum*. The \begin{align*}x-\end{align*}intercepts are (1.67, 0) and (-3.5, 0).

### Explore More

Graph the quadratic equations using a graphing calculator. Find the vertex and \begin{align*}x-\end{align*}intercepts, if there are any. If there are no \begin{align*}x-\end{align*}intercepts, use algebra to find the imaginary solutions. Round all real answers to the nearest hundredth.

- \begin{align*}y=x^2-x-6\end{align*}
- \begin{align*}y=-x^2+3x+28\end{align*}
- \begin{align*}y=2x^2+11x-40\end{align*}
- \begin{align*}y=x^2-6x+7\end{align*}
- \begin{align*}y=x^2+8x+13\end{align*}
- \begin{align*}y=x^2+6x+34\end{align*}
- \begin{align*}y=10x^2-13x-3\end{align*}
- \begin{align*}y=-4x^2+12x-3\end{align*}
- \begin{align*}y=\frac{1}{3}(x-4)^2+12\end{align*}
- \begin{align*}y=-2(x+1)^2-9\end{align*}

**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*}, 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=(x-1)^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^2-5\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.

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 5.17.