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10.1: Graphs of Quadratic Functions

Difficulty Level: At Grade Created by: CK-12

Chapter 9 introduced the concept of factoring quadratic trinomials of the form \begin{align*}0=ax^2+bx+c\end{align*}. This is also called the standard form for a quadratic equation. The most basic quadratic equation is \begin{align*}y=x^2\end{align*}. The word quadratic comes from the Latin word quadrare, meaning “to square.” By creating a table of values and graphing the ordered pairs, you find that a quadratic equation makes a \begin{align*}U\end{align*}-shaped figure called a parabola.

\begin{align*}x\end{align*} \begin{align*}y\end{align*}
–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 \begin{align*}(h, k)\end{align*}.

Because the line of symmetry is a vertical line, its equation has the form \begin{align*}y=h\end{align*}, where \begin{align*}h=\end{align*} the \begin{align*}x-\end{align*}coordinate of the vertex.

An equation of the form \begin{align*}y=ax^2\end{align*} forms a parabola.

If \begin{align*}a\end{align*} is positive, the parabola will open upward. The vertex will be a minimum.

If \begin{align*}a\end{align*} is negative, the parabola will open downward. The vertex will be a maximum.

The variable \begin{align*}a\end{align*} in the equation above is called the leading coefficient of the quadratic equation. Not only will it tell you if the parabola opens up or down, but it will also tell you the width.

If \begin{align*}a>1\end{align*} or \begin{align*}a<-1\end{align*}, the parabola will be narrow about the line of symmetry.

If \begin{align*}-1 < a < 1\end{align*}, the parabola will be wide about the line of symmetry.

Example 1: Determine the direction and shape of the parabola formed by \begin{align*}y=-\frac{1}{2} x^2\end{align*}.

Solution: The value of \begin{align*}a\end{align*} in the quadratic equation is –1.

  • Because \begin{align*}a\end{align*} is negative, the parabola opens downward.
  • Because \begin{align*}a\end{align*} 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 (\begin{align*}x-\end{align*}coordinates).
  • Range is the set of all outputs (\begin{align*}y-\end{align*}coordinates).

The domain of every quadratic equation is all real numbers \begin{align*}(\mathbb{R})\end{align*}. The range of a parabola depends upon whether the parabola opens up or down.

If \begin{align*}a\end{align*} is positive, the range will be \begin{align*}y \ge k\end{align*}.

If \begin{align*}a\end{align*} is negative, the range will be \begin{align*}y \le k\end{align*}, where \begin{align*}k=y-\end{align*}coordinate of the vertex.

Vertical Shifts

Compare the five parabolas to the right. What do you notice?

The five different parabolas are congruent with different \begin{align*}y-\end{align*}intercepts. Each parabola has an equation of the form \begin{align*}y=ax^2+c\end{align*}, where \begin{align*}a=1\end{align*} and \begin{align*}c=y-\end{align*}intercept. In general, the value of \begin{align*}c\end{align*} will tell you where the parabola will intersect the \begin{align*}y-\end{align*}axis.

The equation \begin{align*}y=ax^2+c\end{align*} is a parabola with a \begin{align*}y-\end{align*}intercept of \begin{align*}(0, c)\end{align*}.

The vertical movement along a parabola’s line of symmetry is called a vertical shift.

Example 1: Determine the direction, shape, and \begin{align*}y-\end{align*}intercept of the parabola formed by \begin{align*}y=\frac{3}{2} x^2-4\end{align*}.

Solution: The value of \begin{align*}a\end{align*} in the quadratic equation is \begin{align*}\frac{3}{2}\end{align*}.

  • Because \begin{align*}a\end{align*} is positive, the parabola opens upward.
  • Because \begin{align*}a\end{align*} is greater than 1, the parabola is narrow about its line of symmetry.
  • The value of \begin{align*}c\end{align*} is –4, so the \begin{align*}y-\end{align*}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.

\begin{align*}t\end{align*} \begin{align*}d\end{align*}
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 \begin{align*}a>1\end{align*}, so the graph is narrow about its line of symmetry. However, because the values of the dependent variable \begin{align*}d\end{align*} are very large, the graph is misleading.

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=x-0.04x^2\end{align*}, where \begin{align*}x=\end{align*} distance in feet from the tee and \begin{align*}y=\end{align*} height of the golf ball, in feet.

Describe the shape of this parabola. What is its \begin{align*}y-\end{align*}intercept?

Solution: The value of \begin{align*}a\end{align*} in the quadratic equation is –0.04.

  • Because \begin{align*}a\end{align*} is negative, the parabola opens downward.
  • Because \begin{align*}a\end{align*} is between –1 and 1, the parabola is wide about its line of symmetry.
  • The value of \begin{align*}c\end{align*} is 0, so the \begin{align*}y-\end{align*}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*}. This equation is in standard form \begin{align*}f(x)=ax^2+bx+c\end{align*}, where \begin{align*}a=\frac{1}{20}, b=1\end{align*}, and \begin{align*}c=0\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*}.
  • 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 CK-12 Basic Algebra: Algebra Applications: Quadratic Functions.

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.

CK-12 Basic Algebra: Graphs of Quadratic Functions (16:05)

  1. Define the following terms in your own words.
    1. Vertex
    2. Line of symmetry
    3. Parabola
    4. Minimum
    5. Maximum
  2. Without graphing, how can you tell if \begin{align*}y=ax^2+bx+c\end{align*} opens up or down?
  3. Using the parabola below, identify the following:
    1. Vertex
    2. \begin{align*}y-\end{align*}intercept
    3. \begin{align*}x-\end{align*}intercepts
    4. Domain
    5. Range
    6. Line of symmetry
    7. Is \begin{align*}a\end{align*} positive or negative?
    8. Is \begin{align*}a\end{align*} \begin{align*}-1 < a < 1\end{align*} or \begin{align*}a<-1\end{align*} or \begin{align*}a>1\end{align*}?

  4. Use the stopping distance function from the lesson to find:
    1. \begin{align*}d(45)\end{align*}
    2. What speed has a stopping distance of about 96 feet?
  5. Using Galileo’s law \begin{align*}d=16t^2\end{align*}, find the distance an object has fallen at 3.5 seconds.

Graph the following equations by making a table. Let \begin{align*}-3 \le x \le 3\end{align*}. Determine the range of each equation.

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

Which has a more positive \begin{align*}y-\end{align*}intercept?

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

Determine the direction, shape, and y-intercept of the parabola formed by each equation.

  1. \begin{align*}y=\frac12x^2-2x-8\end{align*}
  2. \begin{align*}y=-3x^2+10x-21\end{align*}
  3. \begin{align*}y=2x^2+6x+4\end{align*}

Determine the direction, shape, and y-intercept of the parabola formed by each equation.

  1. \begin{align*}y=-2x^2-2x-3\end{align*}
  2. \begin{align*}y=3x^2\end{align*}
  3. \begin{align*}y=16-4x^2\end{align*}

Which equation produces a parabola that is narrower around its line of symmetry?

  1. \begin{align*}y=x^2\end{align*} or \begin{align*}y=4x^2\end{align*}
  2. \begin{align*}y=-2x^2\end{align*} or \begin{align*}y=-2x^2 -2\end{align*}
  3. \begin{align*}y=3x^2-3\end{align*} or \begin{align*}y=3x^2-6\end{align*}

Graph the following functions by making a table of values. Use the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts to help you pick values for the table (factor completely and solve for \begin{align*}y=0\end{align*} to find the \begin{align*}x-\end{align*} intercepts.

  1. \begin{align*}y=4x^2-4\end{align*}
  2. \begin{align*}y=-x^2+x+12\end{align*}
  3. \begin{align*}y=2x^2+10x+8\end{align*}
  4. \begin{align*}y=\frac{1}{2} x^2-2x\end{align*}
  5. \begin{align*}y=x-2x^2\end{align*}
  6. \begin{align*}y=4x^2-8x+4\end{align*}
  7. 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+2x-0.16x^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?
  8. 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 x-x^2\end{align*}. From the graph, find what dimensions of the rectangle would give him the greatest area.

Mixed Review

  1. Factor \begin{align*}6u^2 v-11u^2 v^2-10u^2 v^3\end{align*} using its GCF.
  2. Factor into primes: \begin{align*}3x^2+11x+10\end{align*}.
  3. Simplify \begin{align*}- \frac{1}{9} (63) \left(- \frac{3}{7} \right)\end{align*}.
  4. Solve for \begin{align*}b: |b+2|=9\end{align*}.
  5. Simplify \begin{align*}(4x^3 y^2 z)^3\end{align*}.
  6. What is the slope and \begin{align*}y-\end{align*}intercept of \begin{align*}7x+4y=9?\end{align*}

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