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

### Graphing Quadratic Equations

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.

Let's graph \begin{align*}y=-3x^2+14x-8\end{align*}

Using a TI-83/84, 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.

Using a graphing calculator, let's now find the vertex of the parabola above.

To find the vertex, press \begin{align*}2^{nd}\end{align*}**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*}

Using a graphing calculator, let's find the \begin{align*}x-\end{align*}

To find the \begin{align*}x-\end{align*}**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*}

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*}**3:Zoom Out**, ENTER, ENTER.

### Examples

#### Example 1

Earlier, you were asked to find the time at which the arrow reaches its maximum height and to find that height.

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.

#### Example 2

Graph \begin{align*}y=6x^2+11x-35\end{align*}

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*}

### Review

Graph the quadratic equations using a graphing calculator. Find the vertex and \begin{align*}x-\end{align*}

- \begin{align*}y=x^2-x-6\end{align*}
y=x2−x−6 - \begin{align*}y=-x^2+3x+28\end{align*}
y=−x2+3x+28 - \begin{align*}y=2x^2+11x-40\end{align*}
y=2x2+11x−40 - \begin{align*}y=x^2-6x+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^2-13x-3\end{align*}
y=10x2−13x−3 - \begin{align*}y=-4x^2+12x-3\end{align*}
y=−4x2+12x−3 - \begin{align*}y=\frac{1}{3}(x-4)^2+12\end{align*}
y=13(x−4)2+12 - \begin{align*}y=-2(x+1)^2-9\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*}y=12x2 on the same set of axes in the calculator. Describe how \begin{align*}a\end{align*}a effects the shape of the parabola. - Graph \begin{align*}y=x^2, y=-x^2\end{align*}
y=x2,y=−x2 , and \begin{align*}y=-2x^2\end{align*}y=−2x2 on the same set of axes in the calculator. Describe how \begin{align*}a\end{align*}a effects the shape of the parabola. - Graph \begin{align*}y=x^2, y=(x-1)^2\end{align*}
y=x2,y=(x−1)2 , and \begin{align*}y=(x+4)^2\end{align*}y=(x+4)2 on the same set of axes in the calculator. Describe how \begin{align*}h\end{align*}h effects the location of the parabola. - Graph \begin{align*}y=x^2, y=x^2+2\end{align*}
y=x2,y=x2+2 , and \begin{align*}y=x^2-5\end{align*}y=x2−5 on the same set of axes in the calculator. Describe how \begin{align*}k\end{align*}k 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*}y=−0.003x2+1.3x+4 , where \begin{align*}x\end{align*}x is the horizontal distance and \begin{align*}y\end{align*}y is the height (in feet) of the ball. Find the maximum height of the ball and its total distance travelled.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 5.17.