1. Chapter 4 Quadratic Functions and Various Nonlinear Topics Section 4.2

2. Section 4.2 Parabolas and Their Applications
Definition of Quadratic Function
Parabolas: Features and Graphs
General and Vertex Forms of Quadratic Functions
Average Rate of Change

3. A quadratic function is a function of the form f(x) = ax2 + bx + c
Definition of Quadratic Function
A quadratic function is a function of the form
f(x) = ax2 + bx + c
where a, b, and c are real numbers, a  0. 
The graph of a quadratic function resembles a
U-shaped curve and it is called a parabola. It will
have either an upward or a downward concavity.
y = ax2 + bx + c is frequently called the general form of a parabola.

4. In a parabola of the form y = ax2 + bx + c, a  0:
Features of a Parabola
Concavity:
In a parabola of the form y = ax2 + bx + c, a  0:
If a > 0, the parabola is concave up.
if a < 0, it is concave down.
Examples:

5. Vertex: “Turning point" of the graph of the quadratic function.
Features of a Parabola
Vertex: “Turning point" of the graph of the quadratic function.
 If concave up, vertex is the lowest or minimum point.
If concave down, vertex is the highest or maximum
point.
Axis of symmetry: Vertical line through the vertex; divides the parabola into two parts which are mirror images.

6. Vertex of a Parabola of the Form f(x) = ax2 + bx + c, a  0.
x-coordinate:
The x-coordinate of the vertex is given by
y-coordinate:
The y-coordinate of the vertex is given by
Summary:
First, find the x-coordinate of the vertex by using
Then, evaluate the quadratic function at this x-value to find the y-coordinate of the vertex.

7. x-coordinate of the vertex:
Given the quadratic function f(x) = 3x2 – 12x – 15, find the coordinates of the vertex and state whether it is a maximum or minimum point.
x-coordinate of the vertex:
y-coordinate of the vertex:
Coordinates of the vertex: (2, –27)
 a > 0, thus the parabola is concave up and the vertex is a minimum point.

8. If they exist, the x-intercepts of the parabola will occur when y = 0.
Features of a Parabola
X-intercept: The graph of a parabola may cross the x-axis once, or at two different points, or not at all.
If they exist, the x-intercepts of the parabola will occur when y = 0.
So, they are found by setting ax2 + bx + c = 0 and solving for x.
(Solve by factoring, if possible, or applying one of the various solving methods learned in Section 4.1.)

9. Y-intercept: The graph of a parabola crosses the y-axis exactly once.
Features of a Parabola
Y-intercept: The graph of a parabola crosses the
y-axis exactly once.
The y-intercept is found by letting x = 0 in
y = ax2 + bx + c, and solving for y.
Observe when x = 0, we have
y = a(0)2 + b(0) + c = 0
So, in a parabola y = ax2 + bx + c, the y-intercept is given by y = c; in ordered pair form (0, c ).

10. Given the quadratic function y = x2 – 2x – 15, find all intercepts.
y-intercept: Let x = 0 and solve for y.
y = x(0)2 – 2(0) – 15 = –15
In ordered pair form: (0, –15)
Observe the y-intercept is equivalent to the value of c.
x-intercept(s): Let y = 0 and solve for x.
x2 – 2x – 15 = 0
(x + 3)(x – 5) = 0
x = – 3 and x = 5
In ordered pair form: (–3, 0) and (5, 0)

11. Determine the concavity:
Graphing a Parabola of the Form y = ax2 + bx + c, a  0.
Determine the concavity:
If a > 0, the parabola is concave up; if a < 0, concave down
Find the vertex: (–b/2a, f(–b/2a))
If concave up, the vertex is a minimum point
If concave down, the vertex is a maximum point
3. Identify the y-intercept: f(0) is given by the value of c.
4. Find any x-intercepts: Let y = 0 and solve for x.
Plot the vertex, the y-intercept, and x-intercepts (if any),
and connect them with a smooth U-shaped curve. Show the
axis of symmetry. (If needed, use the axis of symmetry to
plot a symmetric point to the y-intercept to complete graph.)

12. a = 1, b = –4 a > 0, therefore, the parabola is concave up.
Graph the quadratic function f(x) = x2 – 4x – 5.
Concavity:
a > 0, therefore, the parabola is concave up.
Vertex: (–b/2a, f(–b/2a))
x-coordinate: –b/2a
a = 1, b = –4
So, x = –(–4)/2(1) = 2
y-coordinate: f(–b/2a)
So, f(2) = (2)2 – 4(2) – 5 = –9
Coordinates of the Vertex: (2, –9)
(continued on the next slide)

13. Graph the quadratic function f(x) = x2 – 4x – 5. Y-intercept: f(0) = c
(Contd.)
Graph the quadratic function f(x) = x2 – 4x – 5.
Y-intercept: f(0) = c
c = –5; y-intercept is (0, –5)
X-intercept(s):
Discriminant, b2 – 4ac = (–4)2 – 4(1)(–5) > 0.
Thus, this parabola will have two zeros (x-intercepts).
x2 – 4x – 5 = 0
(x + 1)(x – 5) = 0
x = –1 and x = 5
x-intercepts are (–1, 0) and (5, 0)
(continued on the next slide)

14. The graph of the quadratic function f(x) = x2 – 4x – 5 is shown below:
(Contd.)
The graph of the quadratic function f(x) = x2 – 4x – 5 is shown below:

15. a > 0, concave up; a < 0, concave down
Vertex Form of a Quadratic Function
f(x) = a(x – h)2 + k, a  0, where (h, k) are the coordinates of the vertex.
a > 0, concave up; a < 0, concave down
If |a| < 1, graph will be wider than the graph of y = x2.
If |a| > 1, graph will be narrower than the graph of y = x2.
The axis of symmetry has the equation x = h.
The h-coordinate of the vertex represents a horizontal shift and it is the “opposite” of the value stated in f(x) = a(x – h)2 + k.
The value of the k-coordinate of the vertex represents a vertical shift, and it will have the same sign as the one in the given function.

16. Given the function f(x) = 3(x – 6)2 + 5, do the following:
a. Determine the vertex and state whether it is a maximum
point or a minimum point.
The coordinates of the vertex are (6, 5). The value of a = 3.
Since a > 0, the graph of the parabola will open upward, which
means that the vertex is a minimum point.
b. Find the equation of the axis of symmetry.
Since h = 6, the equation of the axis of symmetry is x = 6.
c. State whether the graph of the function opens wider or
narrower than the graph of the basic parabola y = x2.
|a| = 3. Narrower.

17. Domain: (–∞, ∞) Range: [5, ∞) (Contd.)
d. Graph the given function, f(x) = 3(x – 6)2 + 5, and
corroborate your answers to parts (a)-(c).
e. State the domain and range of the given function.
Domain: (–∞, ∞)
Range: [5, ∞)

18. Texas Instruments' (TI) revenue (in millions of dollars) from
can be approximated by the function
T(x) = –226.26(x – 5.2) , where x is the number
of years since Sources:
a. Find the vertex of the quadratic function.
(h, k) = (5.2, )
b. Without graphing, determine whether the vertex is a
maximum or minimum point. Verify graphically.
The value of a = – Since a < 0, this parabola is
concave down, thus the vertex is a maximum point.
[0, 15, 1] by [0, 15000, 1000]
(continued on the next slide)

19. Texas Instruments' (TI) revenue (in millions of dollars) from
(Contd.)
Texas Instruments' (TI) revenue (in millions of dollars) from
can be approximated by the function
T(x) = –226.26(x – 5.2) , where x is the number
of years since Sources:
c. Interpret the vertex as it applies to this problem.
The vertex is given by (5.2, ). We know that x is the number of years since 2001, thus x = 5 represents 2006.
In early 2006, the revenue was approximately $
(in millions of dollars).
(continued on the next slide)

20. Texas Instruments' (TI) revenue (in millions of dollars) from
(Contd.)
Texas Instruments' (TI) revenue (in millions of dollars) from
can be approximated by the function
T(x) = –226.26(x – 5.2) , where x is the number
of years since Sources:
d. Use your graphing calculator to graph the function and
find the intercepts graphically. Round your answers to the
nearest tenth.
The x-intercept is (12.8, 0) and the y-intercept is (0, ).
e. Identify the domain and range of the function.
Domain: [0, 12.8] Range: [0, ]

21. Finding the Equation of a Quadratic Function
Given the Vertex and a Point
Write the equation of a parabola whose vertex is (–5, –3) and passes through the point (–4, –1). Graph your equation and verify that the parabola contains the given points.
The vertex of the parabola is at (h, k) = (–5, –3), thus the quadratic function has the form y = a(x + 5)2 – 3.
The parabola passes through (–4, –1). We can substitute these coordinates for x and y and solve for a.
y = a(x + 5)2 – 3
–1 = a(–4 + 5)2 – 3
–1 = a – 3
2 = a
The equation of the parabola is
y = 2(x + 5)2 – 3. See graph next:

22. Finding the Equation of a Quadratic Function Given the Graph
Find the equation of the parabola whose graph is shown.
The vertex of the parabola is at (h, k) = (2, 5).
Using the vertex form, we can substitute any other point on the graph for x and y. We will use the point (3, 2) and solve for a:
y = a(x – 2)2 + 5
2 = a(3 – 2)2 + 5
2 = a + 5
–3 = a
The equation of the parabola is y = –3(x – 2)2 + 5 (vertex form) or, simplifying, y = –3x2 + 12x – 7 (general form).

23. The average rate of change of f between x1 and x2 is given by
Recall the average rate of change of any function is the slope of the secant line between two distinct points on the curve. We can find the average rate of change between two points on the graph of a quadratic function.
Let f be a function. Let (x1, f(x1)) and (x2, f(x2)) represent two distinct points on the graph of f, where x1 ≠ x2.
The average rate of change of f between x1 and x2 is given by

24. The graph of f(x) = 3x2 + 6x – 3 is given below
The graph of f(x) = 3x2 + 6x – 3 is given below. Find the average rate of change between x1 = –2 to x2 = 2.
Using the points (–2, –3) and (2, 21), we calculate the average rate of change:

25. Vertex: (–b/2a, h(–b/2a)) t = –111/2(–16) = 3.5
During a New Year's Eve celebration, fireworks were launched from a platform 7 feet high with an initial velocity of 111 feet per second. The height in feet, h, of the fireworks can be modeled by the function h(t) = –16t t + 7, where t is time in seconds.
a. Find the vertex of the quadratic function. Round your answer to the nearest tenth and interpret using a complete sentence.
Vertex: (–b/2a, h(–b/2a))
t = –111/2(–16) = 3.5
h(3.5) = –16(3.5) (3.5) + 7 = 199.5
Coordinates of the vertex: (3.5, 199.5)
The fireworks reached a maximum height of feet after
3.5 seconds.
(continued on the next slide)

26. b. Find any x-intercept(s) algebraically and graphically. Interpret
(Contd.)
During a New Year's Eve celebration, fireworks were launched from a platform 7 feet high with an initial velocity of 111 feet per second. The height in feet, h, of the fireworks can be modeled by the function h(t) = –16t t + 7, where t is time in seconds.
b. Find any x-intercept(s) algebraically and graphically. Interpret
using a complete sentence.
Let h(t) = 0 and solve the quadratic equation for t.
–16t t + 7 = 0
16t2 – 111t – 7 = (Divide both sides by –1)
(16t + 1)(t – 7) = 0
16t + 1 = or t – 7 = 0
t = –1/ or t = 7
(continued on the next slide)

27. Algebraically, we found that the x-intercepts occur at t = –1/16
(Contd.)
During a New Year's Eve celebration, fireworks were launched from a platform 7 feet high with an initial velocity of 111 feet per second. The height in feet, h, of the fireworks can be modeled by the function h(t) = –16t t + 7, where t is time in seconds.
Algebraically, we found that the x-intercepts occur at t = –1/16
or t = 7.
Next we find the x-intercepts graphically. We will use the
intercepts method (“zero” option of the calculator):
A negative time-value (t = –1/16) makes no sense in the context of this problem. The first remnants of the fireworks reached ground level at t = 7 seconds.
(continued on the next slide)

28. c. Find the vertical intercept and explain its meaning in the
(Contd.)
During a New Year's Eve celebration, fireworks were launched from a platform 7 feet high with an initial velocity of 111 feet per second. The height in feet, h, of the fireworks can be modeled by the function h(t) = –16t t + 7, where t is time in seconds.
c. Find the vertical intercept and explain its meaning in the
context of this problem.
We evaluate h(t) at t = 0:
h(0) = –16(0) (0) + 7 = 7
The vertical intercept is (0, 7). The initial height of the fireworks at 0 seconds was 7 feet.
(continued on the next slide)

29. The increasing interval is (0, 3.5). The decreasing interval is
(Contd.)
d. Find the increasing and decreasing intervals. Interpret these intervals in the context of this problem.
The increasing interval is (0, 3.5). The decreasing interval is
(3.5, 7).
Once launched, the fireworks traveled upward for 3.5 seconds, then started descending towards the ground through the 7th second.
e. Find the domain and range.
Domain: [0, 7]
Range: Recall the coordinates of the vertex are (3.5, 199.5), thus the range is [0, 199.5].

30. Using your textbook, practice the problems assigned by your instructor to review the concepts from Section 4.2.