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Difficulty Level: At Grade Created by: CK-12

## Introduction

The Ball

When passing the baseball field, Mr. Travis handed the students the following problem written on a piece of paper that looked like a baseball. Here is what it said.

When an object is thrown into the air with a starting velocity of r\begin{align*}r\end{align*} feet per second, its distance d\begin{align*}d\end{align*} in feet, above its starting point t\begin{align*}t\end{align*} seconds after it is thrown is about d=rt16t2\begin{align*}d=rt-16t^2\end{align*}. Use your graphing calculator or a table of values and graph paper to show the distance of an object from its starting point that has an initial velocity of 80 feet per second.

To figure out this problem, you will need to know about quadratic functions and their graphs. Pay close attention because you will need to work with this problem again at the end of the lesson.

What You Will Learn

By the end of this lesson, you will be able to execute the following skills.

• Recognize a quadratic function as an equation in two variables that can be written in the form y=ax2+bx+c\begin{align*}y = ax^2 + bx + c\end{align*}, where a\begin{align*}a\end{align*}, b\begin{align*}b\end{align*} and c\begin{align*}c\end{align*} are constants and a0\begin{align*}a \ne 0\end{align*}.
• Evaluate quadratic functions for several values of the domain, using tables.
• Recognize the graph of a quadratic function as a parabola, and associate the value a\begin{align*}a\end{align*} with its upward or downward opening.
• Evaluate and analyze real – world situations modeled by quadratic functions, using a graphing calculator.

Teaching Time

I. Recognize a Quadratic Function as an Equation in Two Variables that Can Be Written in the Standard Form

We have seen many parabolas in the previous section and discussed the relationship between the coefficients in a quadratic equation and its graph. We will take one step further by defining quadratic functions and seeing how they can be applied to the world we live in.

We have touched on the idea of quadratic equations and standard form. You have used input values, oftentimes x\begin{align*}x\end{align*} values, to calculate corresponding y\begin{align*}y\end{align*} values. In those cases, our input values are known as the domain, while the output values our known as the range. These are also called the independent variable and dependent variable, at times.

x\begin{align*}\underline{x}\end{align*}-values y\begin{align*}\underline{y}\end{align*}-values
input values output values
domain range
independent variable dependent variable

You have also learned that a function is a relation that assigns exactly one value of the domain to each value of the range. So, when we say quadratic function, we are referring to any function that can be written in the form y=ax2+bx+c\begin{align*}y=ax^2+bx+c\end{align*}, where a,b,\begin{align*}a, b,\end{align*} and c\begin{align*}c\end{align*} are constants and a0\begin{align*}a \neq 0\end{align*}. This we defined as the standard form.

Why can’t a\begin{align*}a\end{align*} equal zero? What happens if it does? If the a\begin{align*}a\end{align*} value is zero, you might notice that it would make the first term ax2\begin{align*}ax^2\end{align*} disappear because anything times zero is zero. You would be left with simply y=bx+c\begin{align*}y=bx+c\end{align*}. Although this is still a function, it is no longer quadratic. This is a linear function. All quadratic functions are to the 2nd\begin{align*}2^{nd}\end{align*} degree.

Let’s look at some examples.

Tell if the following equations are quadratic functions. If they are, place them in standard form and identify the a,b,\begin{align*}a, b,\end{align*} and c\begin{align*}c\end{align*} values.

1. y=x23x+5\begin{align*}y=x^2-3x+5\end{align*}

Yes, it is a quadratic function. The standard form is y=x23x+5; a=1,b=3,c=5\begin{align*}y=x^2-3x+5; \ a=1,b=-3,c=5\end{align*}.

2. y=7x2+4x\begin{align*}y=-7x^2+4x\end{align*}

Yes, it is a quadratic function. The standard form is y=7x2+4x; a=7,b=4,c=0\begin{align*}y=-7x^2+4x; \ a=-7,b=4,c=0\end{align*}.

3. y6=x2\begin{align*}y-6=x^2\end{align*}

Yes, it is a quadratic function. The standard form is y=x2+6; a=1,b=0,c=6\begin{align*}y=x^2+6; \ a=1,b=0,c=6\end{align*}.

4. 3x2+y=3x2+4x2\begin{align*}3x^2+y=3x^2+4x-2\end{align*}

No, this is not a quadratic function because if you subtract 3x2\begin{align*}3x^2\end{align*} from both sides, your a\begin{align*}a\end{align*} value will be zero.

Write down the definition and form for a quadratic function in your notebook.

II. Evaluate Quadratic Functions for Several Values of the Domain, Using Tables

You know that the word domain refers to input values and the word range refers to output values. Recall that a function is a relation that assigns exactly one value of the domain to each value of the range. That means that for every x\begin{align*}x\end{align*} value, there is only one y\begin{align*}y\end{align*} value.

We can find y\begin{align*}y\end{align*} values by substituting x\begin{align*}x\end{align*} values in the function. We organize the information using a table of values or a t-table. In most cases, the input values could be any numbers. However, for our convenience, we will use negative numbers, zero, and positive numbers as shown in the example below.

Example

Complete a table of values for the function y=x2+3x+2\begin{align*}y=x^2+3x+2\end{align*}.

x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
3\begin{align*}-3\end{align*}
-2
-1
0
1
2
3

To find the y\begin{align*}y\end{align*} values, we will substitute the x\begin{align*}x\end{align*} values in the equation.

The completed t – table should look like this.

x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
3\begin{align*}-3\end{align*} 2
-2 0
-1 0
0 2
1 6
2 12
3 20

Evaluating a quadratic function is always the same. You substitute the x\begin{align*}x\end{align*} – values into the equation and solve the for y\begin{align*}y\end{align*} – values.

III. Recognize the Graph of a Quadratic Function as a Parabola, and Associate the Value a with its Upward or Downward Opening

We defined the standard form of a quadratic function as y=ax2+bx+c\begin{align*}y = ax^2 + bx + c\end{align*}.

You might remember that we discussed the effect that the values of \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*} have on the graphs of quadratic equations. Now we are going to use this information when we look at a quadratic function. What we know about the values of \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*} help us to understand the opening of a parabola.

Value What it tells you Example \begin{align*}y = -3x^2 + x -2\end{align*}
\begin{align*}a\end{align*}

if \begin{align*}a > 0\end{align*}, graph opens upward

if \begin{align*}a < \end{align*} graph opens downward

if \begin{align*}a\end{align*} is close to zero, wider graph

if \begin{align*}a\end{align*} is far from zero, narrower graph

\begin{align*}a = -3\end{align*}

\begin{align*}a\end{align*} is less than zero so graph opens downward

\begin{align*}a\end{align*} is further from zero so will be narrow

\begin{align*}b\end{align*} helps predict axis of symmetry

\begin{align*}b = 1\end{align*}

axis of symmetry of the parabola

\begin{align*}c\end{align*} \begin{align*}y\end{align*}-intercept

\begin{align*}c = -2\end{align*}

graph crosses \begin{align*}y\end{align*}-axis at -2

We know that the graph of a quadratic function will always be a parabola. A parabola is a kind of “U” shape that is always symmetrical on both sides. It can go either upward or downward. Also, a parabola is not linear—no part of the parabola is actually a straight line. Thus, it cannot be vertical, either.

If we wanted to predict the shape of the parabola, we would need to look at the value of \begin{align*}a\end{align*}. We know that \begin{align*}a\end{align*} helps us to determine a parabola’s shape.

Look at these examples.

Examples

\begin{align*}& y = x^2 && y = 3x^2 && y = \frac{1}{3}x^2\\ & a = 1 \ so \ a > 0 && a = 3 \ so \ a > 0 && a = \frac{1}{3} \ so \ a > 0\\ & \text{graph opens upward} && \text{graph opens upward} && \text{graph opens upward}\\ & \text{neither wide nor narrow} && \text{graph is narrow} && \text{graph is wide}\end{align*}

Now that you understand how these graphs look and how the equation of the graph affects its appearance, it is time to make some predictions.

Example

What would you predict about the graph of \begin{align*}y = 7x^2\end{align*}?

Because the \begin{align*}a\end{align*} value is 7, it would be very narrow. Also, because \begin{align*}a > 0\end{align*}, it would open upward.

Example

What would you predict about the graph of \begin{align*}y = - \frac{1}{4}x^2\end{align*}?

Because the \begin{align*}a\end{align*} value is \begin{align*}-\frac{1}{4}\end{align*}, it would be very wide. Also, because \begin{align*}a < 0\end{align*}, it would open downward.

IV. Evaluate and Analyze Real – World Situations Modeled by Quadratic Functions, Using a Graphing Calculator

Einstein’s famous Theory of Relativity is a quadratic function! He knew that there were many applications to the of quadratic equations to the real world.

There are many quadratic functions and formulas used in every day life that you can use by making a table of values or using a graphing calculator. As we do so, let’s keep in mind which values for the domain are realistic.

Example

The area of a square is found with the formula \begin{align*}A=s^2\end{align*}. Show a graph of the area as the length of the side increases.

\begin{align*}& \text{Length of side} \quad 0 \qquad 1 \qquad 2 \qquad 3 \qquad 4 \qquad 5\\ & \text{Area} \qquad \qquad \quad 0 \qquad 1 \qquad 4 \qquad 9 \qquad 16 \quad \ 25\end{align*}

The area of a square is non-linear, so the graph will be non-linear too. Let’s look at the graph that represents the area of a square.

You can see in this graph that the area of a square is not a linear equation. Rather, it is parabolic.

In this graph, it is necessary to show only quadrant I because distance and area cannot be negative.

You can graph this function in your graphing calculator.

Now let’s apply what we have learned to the problem from the introduction.

## Real-Life Example Completed

The Ball

Here is the problem from the introduction. Reread it and then solve it with your graphing calculator.

When passing the baseball field, Mr. Travis handed the students the following problem written on a piece of paper that looked like a baseball. Here is what it said.

When an object is thrown into the air with a starting velocity of \begin{align*}r\end{align*} feet per second, its distance \begin{align*}d\end{align*} in feet, above its starting point \begin{align*}t\end{align*} seconds after it is thrown is about \begin{align*}d=rt-16t^2\end{align*}. Use your graphing calculator or a table of values and graph paper to show the distance of an object from its starting point that has an initial velocity of 80 feet per second.

Remember to use your graphing calculator to solve this problem.

Solution to Real – Life Example

First, think about the information that you have and the equation that you can write.

\begin{align*}r & = 80\\ d & = 80t - 16t^2\end{align*}

Next, we can make a table of values.

\begin{align*}& t \ (seconds) \quad \qquad 0 \qquad 1 \qquad 2 \qquad 3 \qquad 4 \qquad 5\\ & d \text{ distance } (ft) \quad \ 0 \quad \ \ 64 \quad \ \ 96 \quad \ 96 \quad \ \ 64 \quad \ \ 0\end{align*}

Finally we can take those values, insert them into a graphing calculator and create the following graph.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Domain
input value, independent value
Range
output value, dependent value
Function
relation that assigns one value of the domain to each value of range
To the \begin{align*}2^{nd}\end{align*} degree in standard form-a parabola is created by a quadratic function.

## Time to Practice

Directions: Tell if the following equations are quadratic functions. If they are, place them in standard form and identify the \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*} values.

1. \begin{align*}y = 3x^2 - x + 4\end{align*}
2. \begin{align*}y = 4x - 18 + x^2\end{align*}
3. \begin{align*}y + x^2 - 6 = x^2 + 4x - 5\end{align*}
4. \begin{align*}3y + 3 = 7x^2 - 5x\end{align*}
5. \begin{align*}6y = x^5 + 3x^4 + x - 7\end{align*}
6. \begin{align*}4y + 3x = 8x^2 + 3x - 12\end{align*}

Directions: Use a table of values to evaluate the following quadratic function for several values in the domain.

1. \begin{align*}y = x^2 - 8\end{align*}
2. \begin{align*}y = - 2x^2 + 5x\end{align*}
3. \begin{align*}y = -x^2 + 3x - 7\end{align*}
4. \begin{align*}y = \frac{2}{3}x^2 + 2x - 1\end{align*}

Use your tables to graph the four functions. Then relate their \begin{align*}a\end{align*} values to the shape of their graphs.

1. \begin{align*}y = x^2 - 8\end{align*}
2. \begin{align*}y = - 2x^2 + 5x\end{align*}
3. \begin{align*}y = -x^2 + 3x - 7\end{align*}
4. \begin{align*}y = \frac{2}{3}x^2 + 2x - 1\end{align*}
5. The formula for the area of a circle is \begin{align*}A=\pi r^2\end{align*}. Use 3.14 for \begin{align*}\pi\end{align*}. Use your graphing calculator or create a table of values to show the area of circles with radii of 0 to 10.
6. When an object is thrown into the air with a starting velocity of \begin{align*}r\end{align*} feet per second, its distance \begin{align*}d\end{align*} in feet, above its starting point \begin{align*}t\end{align*} seconds after it is thrown is about \begin{align*}d= rt - 16t^2\end{align*}. Use your graphing calculator or a table of values and graph paper to show the distance of an object from its starting point that has an initial velocity of 112 feet per second.

17 – 20 Write three equations that are considered quadratic equations and then graph them.

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