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
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
Why can’t
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
We can find
Complete a table of values for the function
















To find the
The completed t – table should look like this.

















Evaluating a quadratic function is always the same. You substitute the
The values of
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*}
if \begin{align*}a < \end{align*}
if \begin{align*}a\end{align*}
if \begin{align*}a\end{align*} 
\begin{align*}a = 3\end{align*}
\begin{align*}a\end{align*}
\begin{align*}a\end{align*} 
\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*} 
\begin{align*}c = 2\end{align*}
graph crosses \begin{align*}y\end{align*} 
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*}
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*}
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*}
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.
Vocabulary
 Domain
 input value, independent value
 Range
 output value, dependent value
 Function
 relation that assigns one value of the domain to each value of range
 Quadratic Function

To the \begin{align*}2^{nd}\end{align*}
2nd degree in standard forma parabola is created by a quadratic function.
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*}
Video Review
Practice
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*}3y1 = 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*}