Have you ever thought about the speed of a baseball? Take a look at this dilemma.
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 a table of values to show the distance of an object from its starting point that has an initial velocity of 80 feet per second. Then graph the velocity of the ball.
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 Concept.
Guidance
A function is a relation that assigns exactly one value of the domain to each value of the range. When we say quadratic function , we are referring to any function that can be written in the form @$\begin{align*}y=ax^2+bx+c\end{align*}@$ , where @$\begin{align*}a, b,\end{align*}@$ and @$\begin{align*}c\end{align*}@$ are constants and @$\begin{align*}a \neq 0\end{align*}@$ . This we defined as the standard form.
Why can’t @$\begin{align*}a\end{align*}@$ equal zero? What happens if it does? If the @$\begin{align*}a\end{align*}@$ value is zero, you might notice that it would make the first term @$\begin{align*}ax^2\end{align*}@$ disappear because anything times zero is zero. You would be left with simply @$\begin{align*}y=bx+c\end{align*}@$ . Although this is still a function, it is no longer quadratic. This is a linear function. All quadratic functions are to the @$\begin{align*}2^{nd}\end{align*}@$ degree.
Let's look at quadratic functions in more detail.
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 @$\begin{align*}x\end{align*}@$ value, there is only one @$\begin{align*}y\end{align*}@$ value.
We can find @$\begin{align*}y\end{align*}@$ values by substituting @$\begin{align*}x\end{align*}@$ 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.
Complete a table of values for the function @$\begin{align*}y=x^2+3x+2\end{align*}@$ .
@$\begin{align*}x\end{align*}@$ | @$\begin{align*}y\end{align*}@$ |
---|---|
@$\begin{align*}-3\end{align*}@$ | |
@$\begin{align*} -2\end{align*}@$ | |
@$\begin{align*} -1\end{align*}@$ | |
@$\begin{align*}{\color{white}-} 0\end{align*}@$ | |
@$\begin{align*}{\color{white}-} 1\end{align*}@$ | |
@$\begin{align*}{\color{white}-} 2\end{align*}@$ | |
@$\begin{align*}{\color{white}-} 3\end{align*}@$ |
To find the @$\begin{align*}y\end{align*}@$ values, we will substitute the @$\begin{align*}x\end{align*}@$ values in the equation.
The completed t – table should look like this.
@$\begin{align*}x\end{align*}@$ | @$\begin{align*}y\end{align*}@$ |
---|---|
@$\begin{align*}-3\end{align*}@$ | @$\begin{align*} 2\end{align*}@$ |
@$\begin{align*} -2\end{align*}@$ | @$\begin{align*} 0\end{align*}@$ |
@$\begin{align*} -1\end{align*}@$ | @$\begin{align*} 0\end{align*}@$ |
@$\begin{align*}{\color{white}-} 0\end{align*}@$ | @$\begin{align*} 2\end{align*}@$ |
@$\begin{align*}{\color{white}-} 1\end{align*}@$ | @$\begin{align*} 6\end{align*}@$ |
@$\begin{align*}{\color{white}-} 2\end{align*}@$ | @$\begin{align*} 12\end{align*}@$ |
@$\begin{align*}{\color{white}-} 3\end{align*}@$ | @$\begin{align*} 20\end{align*}@$ |
Evaluating a quadratic function is always the same. You substitute the @$\begin{align*}x\end{align*}@$ – values into the equation and solve the for @$\begin{align*}y\end{align*}@$ – values.
The values of @$\begin{align*}a, b,\end{align*}@$ and @$\begin{align*}c\end{align*}@$ have an effect 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.
Take a look.
@$$\begin{align*}& y = x^2 && y = 3x^2 && y = \frac{1}{3}x^2\\ & a = 1 \ so \ a > 0 && a = 3 \ so \ a > 0 && a = \frac{1}{3} \ so \ a > 0\\ & \text{graph opens upward} && \text{graph opens upward} && \text{graph opens upward}\\ & \text{neither wide nor narrow} && \text{graph is narrow} && \text{graph is wide}\end{align*}@$$
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.
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.
Answer the following questions by making predictions.
Example A
Predict the opening of @$\begin{align*}-3x^2+4\end{align*}@$ .
Solution: It will open downwards because the @$\begin{align*}a\end{align*}@$ value is negative.
Example B
For the quadratic function in Example A, where will the vertex be?
Solution: At @$\begin{align*}4\end{align*}@$
Example C
Which graph will have a wider opening one with a vertex at 0 or one with a vertex at 8?
Solution: Vertex at 0
Now let's go back to the dilemma from the beginning of the Concept.
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.
Guided Practice
Here is one for you to try on your own.
What would you predict about the graph of @$\begin{align*}y = - \frac{1}{4}x^2\end{align*}@$ ?
Solution
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.
Video Review
Explore More
Use your tables to graph the following functions.
- @$\begin{align*}y = x^2 - 8\end{align*}@$
- @$\begin{align*}y = 3x^2 - x + 4\end{align*}@$
- @$\begin{align*}y = 2x^2 + 4\end{align*}@$
- @$\begin{align*}2y = 4x^2 + 4\end{align*}@$
- @$\begin{align*}3y = 6x^2 + 12\end{align*}@$
- @$\begin{align*}4y = 2x^2 - 12\end{align*}@$
- @$\begin{align*}3y-1 = 6x^2 + 11\end{align*}@$
- @$\begin{align*}2y+2 = 2x^2 + 4\end{align*}@$
- @$\begin{align*}y = - 2x^2 + 5x\end{align*}@$
- @$\begin{align*}y = -x^2 + 3x - 7\end{align*}@$
- @$\begin{align*}y = \frac{2}{3}x^2 + 2x - 1\end{align*}@$
- @$\begin{align*}y = x^2 + 8\end{align*}@$
- @$\begin{align*}y = - 2x^2 + 5x\end{align*}@$
- @$\begin{align*}y = -x^2 + 3x - 1\end{align*}@$
- @$\begin{align*}y = 3x^2 + 2x + 1\end{align*}@$