10.1: Graphs of Quadratic Functions
Chapter 9 introduced the concept of factoring quadratic trinomials of the form



–2  4 
–1  1 
0  0 
1  1 
2  4 
The Anatomy of a Parabola
A parabola can be divided in half by a vertical line. Because of this, parabolas have symmetry. The vertical line dividing the parabola into two equal portions is called the line of symmetry. All parabolas have a vertex, the ordered pair that represents the bottom (or the top) of the curve.
The vertex of a parabola has an ordered pair
Because the line of symmetry is a vertical line, its equation has the form
An equation of the form
If
If
The variable
If
If
Example 1: Determine the direction and shape of the parabola formed by
Solution: The value of
 Because
a is negative, the parabola opens downward.  Because
a is between –1 and 1, the parabola is wide about its line of symmetry.
Domain and Range
Several times throughout this textbook, you have experienced the terms domain and range. Remember:
 Domain is the set of all inputs (
x− coordinates).  Range is the set of all outputs (
y− coordinates).
The domain of every quadratic equation is all real numbers
If
If
Vertical Shifts
Compare the five parabolas to the right. What do you notice?
The five different parabolas are congruent with different
The equation
The vertical movement along a parabola’s line of symmetry is called a vertical shift.
Example 1: Determine the direction, shape, and
Solution: The value of
 Because
a is positive, the parabola opens upward.  Because
a is greater than 1, the parabola is narrow about its line of symmetry.  The value of
c is –4, so they− intercept is (0, –4).
Projectiles are often described by quadratic equations. When an object is dropped from a tall building or cliff, it does not travel at a constant speed. The longer it travels, the faster it goes. Galileo described this relationship between distance fallen and time. It is known as his kinematical law. It states the “distance traveled varies directly with the square of time.” As an algebraic equation, this law is:
Use this information to graph the distance an object travels during the first six seconds.



0  0 
1  16 
2  64 
3  144 
4  256 
5  400 
6  576 
The parabola opens upward and its vertex is located at the origin. The value of
Example 2: Anne is playing golf. On the fourth tee, she hits a slow shot down the level fairway. The ball follows a parabolic path described by the equation, \begin{align*}y=x0.04x^2\end{align*}
Describe the shape of this parabola. What is its \begin{align*}y\end{align*}
Solution: The value of \begin{align*}a\end{align*}
 Because \begin{align*}a\end{align*}
a is negative, the parabola opens downward.  Because \begin{align*}a\end{align*}
a is between –1 and 1, the parabola is wide about its line of symmetry.  The value of \begin{align*}c\end{align*}
c is 0, so the \begin{align*}y\end{align*}y− intercept is (0, 0).
The distance it takes a car to stop (in feet) given its speed (in miles per hour) is given by the function \begin{align*}d(s)=\frac{1}{20} \ s^2+s\end{align*}
Graph the function by making a table of speed values.
\begin{align*}s\end{align*} 
\begin{align*}d\end{align*} 

0  0 
10  15 
20  40 
30  75 
40  120 
50  175 
60  240 
 The parabola opens upward with a vertex at (0, 0).
 The line of symmetry is \begin{align*}x=0\end{align*}
x=0 .  The parabola is wide about its line of symmetry.
Using the function to find the stopping distance of a car travelling 65 miles per hour yields:
\begin{align*}d(65)=\frac{1}{20} (65)^2+65=276.25 \ feet\end{align*}
Multimedia Link: For more information regarding stopping distance, watch this CK12 Basic Algebra: Algebra Applications: Quadratic Functions
 YouTube video.
Practice Set
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.
CK12 Basic Algebra: Graphs of Quadratic Functions (16:05)
 Define the following terms in your own words.
 Vertex
 Line of symmetry
 Parabola
 Minimum
 Maximum
 Without graphing, how can you tell if \begin{align*}y=ax^2+bx+c\end{align*}
y=ax2+bx+c opens up or down?  Using the parabola below, identify the following:
 Vertex

\begin{align*}y\end{align*}
y− intercept 
\begin{align*}x\end{align*}
x− intercepts  Domain
 Range
 Line of symmetry
 Is \begin{align*}a\end{align*}
a positive or negative?  Is \begin{align*}a\end{align*}
a \begin{align*}1<a<1\end{align*}−1<a<1 or \begin{align*}a<1\end{align*}a<−1 or \begin{align*}a>1\end{align*}a>1 ?
 Use the stopping distance function from the lesson to find:

\begin{align*}d(45)\end{align*}
d(45)  What speed has a stopping distance of about 96 feet?

\begin{align*}d(45)\end{align*}
 Using Galileo’s law from the lesson, find:
 The distance an object has fallen at 3.5 seconds
 The total distance the object has fallen in 3.5 seconds
Graph the following equations by making a table. Let \begin{align*}3 \le x \le 3\end{align*}

\begin{align*}y=2x^2\end{align*}
y=2x2 
\begin{align*}y=x^2\end{align*}
y=−x2 
\begin{align*}y=x^22x+3\end{align*}
y=x2−2x+3 
\begin{align*}y=2x^2+4x+1\end{align*}
y=2x2+4x+1 
\begin{align*}y=x^2+3\end{align*}
y=−x2+3 
\begin{align*}y=x^28x+3\end{align*}
y=x2−8x+3 
\begin{align*}y=x^24\end{align*}
y=x2−4
Which has a more positive \begin{align*}y\end{align*}

\begin{align*}y=x^2\end{align*}
y=x2 or \begin{align*}y=4x^2\end{align*}y=4x2 
\begin{align*}y=2x^2+4\end{align*}
y=2x2+4 or \begin{align*}y=\frac{1}{2} x^2+4\end{align*}y=12x2+4  \begin{align*}y=2x^22\end{align*} or \begin{align*}y=x^22\end{align*}
Identify the vertex and \begin{align*}y\end{align*}intercept. Is the vertex a maximum or a minimum?
 \begin{align*}y=x^22x8\end{align*}
 \begin{align*}y=x^2+10x21\end{align*}
 \begin{align*}y=2x^2+6x+4\end{align*}
Does the graph of the parabola open up or down?
 \begin{align*}y=2x^22x3\end{align*}
 \begin{align*}y=3x^2\end{align*}
 \begin{align*}y=164x^2\end{align*}
Which equation has a larger vertex?
 \begin{align*}y=x^2\end{align*} or \begin{align*}y=4x^2\end{align*}
 \begin{align*}y=2x^2\end{align*} or \begin{align*}y=2x^2 2\end{align*}
 \begin{align*}y=3x^23\end{align*} or \begin{align*}y=3x^26\end{align*}
Graph the following functions by making a table of values. Use the vertex and \begin{align*}x\end{align*}intercepts to help you pick values for the table.
 \begin{align*}y=4x^24\end{align*}
 \begin{align*}y=x^2+x+12\end{align*}
 \begin{align*}y=2x^2+10x+8\end{align*}
 \begin{align*}y=\frac{1}{2} x^22x\end{align*}
 \begin{align*}y=x2x^2\end{align*}
 \begin{align*}y=4x^28x+4\end{align*}
 Nadia is throwing a ball to Peter. Peter does not catch the ball and it hits the ground. The graph shows the path of the ball as it flies through the air. The equation that describes the path of the ball is \begin{align*}y=4+2x0.16x^2\end{align*}. Here, \begin{align*}y\end{align*} is the height of the ball and \begin{align*}x\end{align*} is the horizontal distance from Nadia. Both distances are measured in feet. How far from Nadia does the ball hit the ground? At what distance, \begin{align*}x\end{align*}, from Nadia, does the ball attain its maximum height? What is the maximum height?
 Peter wants to enclose a vegetable patch with 120 feet of fencing. He wants to put the vegetable patch against an existing wall, so he needs fence for only three of the sides. The equation for the area is given by \begin{align*}a=120 xx^2\end{align*}. From the graph, find what dimensions of the rectangle would give him the greatest area.
Mixed Review
 Factor \begin{align*}6u^2 v11u^2 v^210u^2 v^3\end{align*} using its GCF.
 Factor into primes: \begin{align*}3x^2+11x+10\end{align*}.
 Simplify \begin{align*} \frac{1}{9} (63) \left( \frac{3}{7} \right)\end{align*}.
 Solve for \begin{align*}b: b+2=9\end{align*}.
 Simplify \begin{align*}(4x^3 y^2 z)^3\end{align*}.
 What is the slope and \begin{align*}y\end{align*}intercept of \begin{align*}7x+4y=9?\end{align*}