§ 8.3
DESCRIPTION
§ 8.3. Quadratic Functions and Their Graphs. Graphing Quadratic Functions. The graph of any quadratic function is a parabola . . Whether the parabola opens upward or downward depends on the coefficient a of the leading term of the quadratic. - PowerPoint PPT PresentationTRANSCRIPT
§ 8.3
Quadratic Functions and Their Graphs
Graphing Quadratic Functions
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 8.3
The graph of any quadratic function is a parabola.
cbxaxy 2
Whether the parabola opens upward or downward depends on the coefficient a of the leading term of the quadratic.• If a is positive, the parabola opens upward (like a bowl).• If a is negative, the parabola opens downward (like an inverted bowl).
Graphs of Quadratic FunctionsThe graph of the quadratic function
is called a parabola. 0,2 acbxaxxf
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 8.3
Graphing Quadratic Functions
-5
0
5
10
15
20
25
30
-6 -4 -2 0 2 4 6-6
0
6
12
18
24
30
-6 -4 -2 0 2 4 6
-6
0
6
12
18
24
30
-6 -4 -2 0 2 4 6
2xxf 23xxf
2
31 xxf
Standard
Narrow
Wide
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 8.3
Graphing Quadratic Functions
Graphing Quadratic Functions With Equations in the Form T
To graph 1) Determine whether the parabola opens upward or downward. If a > 0, it opens upward. It a < 0, it opens downward.
2) Determine the vertex of the parabola. The vertex is (h, k).3) Find any x-intercepts by replacing f (x) with 0. Solve the resulting Quadratic equation for x.
4) Find the y-intercept by replacing x with 0.5) Plot the intercepts and vertex and additional points as necessary. Connect these points with a smooth curve that is shaped like a cup.
,2 khxaxf
khxaxf 2
Page 592
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 8.3
Graphing Quadratic FunctionsEXAMPLE
Graph the functionSOLUTION
We can graph this function by following the steps in the preceding box. We begin by identifying values for a, h, and k.
.842 2 xxf
khxaxf 2
842 2 xxf
a = -2
b = -4
c = -8
1) Determine how the parabola opens. Note that a, the coefficient of , is -2. Thus, a < 0; this negative value tells us that the parabola opens downward.
2x
Graphing Quadratic Functions
.842 2 xxf
2) Find the vertex. The vertex of the parabola is at (h, k). Because h = -4 and k = -8, the parabola has its vertex at (-4, -8).
CONTINUED
3) Find the x-intercepts. Replace f (x) with 0 in
8420 2 x Find x-intercepts, setting f (x) equal to 0.
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 8.3
842 2 x Add to both sides. 242 x 44 2 x Divide both sides by 2.
44 x Apply the square root property.ix 24 Simplify the radical.
ix 24 Subtract 4 from both sides.
Graphing Quadratic Functions
.842 2 xxf
Since no real solutions resulted from this step, there are no x-intercepts.
CONTINUED
4) Find the y-intercept. Replace x with 0 in
40832816284284020 22 f
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 8.3
The y-intercept is -40. The parabola passes through (0,-40).
5) Graph the parabola. With a vertex at (-4,-8), no x-intercepts, and a y-intercept at -40, the graph of f is shown below. The axis of symmetry is the vertical line whose equation is x = -4.
Graphing Quadratic FunctionsCONTINUE
D
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 8.3
-80
-70
-60
-50
-40
-30
-20
-10
0-15 -10 -5 0 5
y-intercept is: -40
Vertex: (-4,-8)
Axis of symmetry: x = -4
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 8.3
Graphing Quadratic FunctionsEXAMPLE Problems from homework
Find the vertex
10. .1223 2 xxf
khxaxf 2
842 2 xxf12.
12,2
8,4
kh,
Find the vertex, intercepts, and sketch
21 2 xxf18.
23 2 xxf20.
41)( 2 xxfCheck point 1
Graphing Quadratic Functions
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 8.3
The Vertex of a Parabola Whose Equation is T
Consider the parabola defined by the quadratic function
The parabola’s vertex is
cbxaxxf 2
.2 cbxaxxf
.2
,2
abf
ab
Page 594.
44,
2
2
abac
ab
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 8.3
Graphing Quadratic FunctionsEXAMPLE
Graph the function Use the graph to identify its domain and its range.
SOLUTION
1) Determine how the parabola opens. Note that a, the coefficient of , is 1. Thus, a > 0; this positive value tells us that the parabola opens upward.
.46 2xxxf
2) Find the vertex. We know that the x-coordinate of the vertex
is We identify a, b, and c for the given function.
Note that a = 1, b = -4, and c = 6.
.2abx
2x
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 8.3
Graphing Quadratic Functions
.248622462 2 f
Substitute the values of a and b into the equation for the x-coordinate:
.2224
124
2
a
bx
CONTINUED
The x-coordinate for the vertex is 2. We substitute 2 for x in the equation of the function to find the corresponding y-coordinate.
The vertex is (2,2).3) Find the x-intercepts. Replace f (x) with 0 in the original function. We obtain . This equation cannot be solved by factoring. We will use the quadratic formula instead.
2460 xx
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 8.3
Graphing Quadratic FunctionsCONTINUE
D
Clearly, the discriminant is going to be negative, 16 – 24 = -8. Therefore, there will be no x-intercepts for the graph of the function.
600600460 2 f
2
2416412
614442
4 22
a
acbbx
4) Find the y-intercept. Replace x with 0 in the original function.
The y-intercept is 6. The parabola passes through (0,6).
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 8.3
Graphing Quadratic FunctionsCONTINUE
D 5) Graph the parabola. With a vertex of (2,2), no x-intercepts, and a y-intercept at 6, the graph of f is shown below. The axis of symmetry is the vertical line whose equation is x = 2.
0
5
10
15
20
25
30
-4 -2 0 2 4 6 8
Vertex: (2,2)
y-intercept is: 6
0
5
10
15
20
25
30
-4 -2 0 2 4 6 8
Axis of symmetry: x = 2
Domain: All real numbers
Range: All real numbers greater than or equal to 2.
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 8.3
Graphing Quadratic FunctionsCONTINUE
D Now we are ready to determine the domain and range of the original function. We can use the second parabola from the preceding page to do so. To find the domain, look for all inputs on the x-axis that correspond to points on the graph.
. ,or number real a is | is ofDomain xxf
To find the range, look for all the outputs on the y-axis that correspond to points on the graph. Looking at the first parabola from the preceding page, we see the parabola’s vertex is (2,2). This is the lowest point on the graph. Because the y-coordinate of the vertex is 2, outputs on the y-axis fall at or above 2.
. 2,or 2| is of Range yyf
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 8.3
Graphing Quadratic FunctionsEXAMPLE Problems
Find the vertex14. 1123 2 xxxf
11,2
.2 cbxaxxf .
2,
2
abf
ab
.4
4,2
2
abac
ab
2
34)12(134
1112
14412
11,2
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 8.3
Graphing Quadratic FunctionsEXAMPLE Problems
Find the vertex, intercepts, and sketch
1522 xxxf28. 245 xxxf 32.
.2 cbxaxxf .2
,2
abf
ab
.4
4,2
2
abac
ab
142 xxxf34.
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 8.3
Minimums & Maximums
Minimum and Maximum: Quadratic FunctionsConsider
1) If a > 0, then f has a minimum that occurs at
This minimum value is
2) If a < 0, then f has a maximum that occurs at
This maximum value is
.2 cbxaxxf
.2abx
.2abx
.2
abf
.2
abf
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 8.3
Minimums & MaximumsEXAMPLE (similar to number 59 and
Example 6)A person standing close to the edge on the top of a 200-foot building throws a baseball vertically upward. The quadratic function
models the ball’s height above the ground, s (t), in feet, t seconds after it was thrown.
How many seconds does it take until the ball finally hits the ground? Round to the nearest tenth of a second.
2006416 2 ttts
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 8.3
Minimums & Maximums
SOLUTIONCONTINUE
D
It might first be useful to have some sort of picture representing the situation. Below is some sort of picture.
0
50
100
150
200
250
300
0 2 4 6 8
Time (seconds)
Heig
th o
f Bal
l (fe
et)
2006416 2 ttts
Point of interest
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 8.3
Minimums & MaximumsCONTINUE
D When the ball is released, it is at a height of 200 feet. That is, when the ball is released, the value of s (t) = 200. By the same token, when the ball finally hits the ground, it will of course be 0 feet above the ground. That is, when the ball hits the ground, the value of s (t) = 0. Therefore, to determine for what value of t the ball hits the ground, we replace s (t) with 0 in the original function.
2006416 2 ttts This is the given function.20064160 2 tt Replace s (t) with 0.
258280 2 tt Factor -8 out of all terms.25820 2 tt Divide both sides by -8.
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 8.3
Minimums & MaximumsCONTINUE
D We will use the quadratic formula to solve this equation.
4
20064822
2524882
4 22
a
acbbt
42.168or
42.168
42.168
42648
42.8or
42.24
05.2or 6.05
Since time cannot be a negative quantity, the answer cannot be -2.05 seconds. Therefore, the ball hits the ground after 6.05 seconds (to the nearest tenth of a second).
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 8.3
Minimums & Maximums (handout included)
EXAMPLE (Number 59 similar to problem above and Example 6)A person standing close to the edge on the top of a 160-foot building throws a
baseball vertically upward. The quadratic function
models the ball’s height above the ground, s (t), in feet, t seconds after it was thrown.
a. After how many seconds does the ball reach its maximum height? What is the maximum height.
b. How many seconds does it take until he ball finally hits the ground? Round to the nearest tenth of a second.
c. Find s(0) and describe what this means.
d. Use your results from parts(a) through (c) to graph the quadratic function. Begin the graph with t=0 and end with the value of t for which the ball hits the ground.
1606416 2 ttts
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 8.3
Minimums & MaximumsStrategy for Solving Problems Involving Maximizing
or Minimizing Quadratic Functions1) Read the problem carefully and decide which quantity is to be maximized or minimized.
2) Use the conditions of the problem to express the quantity as a function in one variable.
3) Rewrite the function in the form4) Calculate . If a > 0, f has a minimum at . This minimum
value is . If a < 0, f has a maximum at . This maximum
value is .
5) Answer the question posed in the problem.
.2 cbxaxxf
abx2
abf
2
ab2
abx2
abf
2
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 8.3
Minimums & MaximumsEXAMPLE (similar to 61 and Example
7)Among all pairs of numbers whose sum is 20, find a pair whose product is as large as possible. What is the maximum product?
SOLUTION1) Decide what must be maximized or minimized. We must
maximize the product of two numbers. Calling the numbers x and y, and calling the product P, we must maximize
P = xy.2) Express this quantity as a function in one variable. In the formula P = xy, P is expressed in terms of two variables, x and y. However, because the sum of the numbers is 20, we can write
x + y = 20.
Blitzer, Intermediate Algebra, 5e – Slide #26 Section 8.3
Minimums & Maximums
We can solve this equation for y in terms of x, substitute the result into P = xy, and obtain P as a function of one variable.
CONTINUED
y = 20 - x Subtract x from both sides of the equation: x + y = 20.
Now we substitute 20 – x for y in P = xy.P = xy = x(20 – x).
Because P is now a function of x, we can writeP (x) = x(20 – x).
3) Write the function in the form . We apply the distributive property to obtain
cbxaxxf 2
Blitzer, Intermediate Algebra, 5e – Slide #27 Section 8.3
Minimums & MaximumsCONTINUE
D
4) Calculate . If a < 0, the function has a maximum at
this value. The voice balloons show that a = -1 and b = 20.
P (x) = (20 – x)x = 20x - .2x
b = 20 a = -1
ab
2
10102
2012
202
a
bx
This means that the product, P, of two numbers who sum is 20 is a maximum when one of the numbers, x, is 10.
Blitzer, Intermediate Algebra, 5e – Slide #28 Section 8.3
Minimums & MaximumsCONTINUE
D 5) Answer the question posed by the problem. The problem asks for the two numbers and the maximum product. We found that one of the numbers, x, is 10. Now we must find the second number, y.
The number pair whose sum is 20 and whose product is as large as possible is 10, 10. The maximum product is 10 x 10 = 100.
y = 20 – x = 20 – 10 = 10.
Blitzer, Intermediate Algebra, 5e – Slide #29 Section 8.3
Minimums & MaximumsEXAMPLE (number 65 and example 8)
You have 200 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed?
SOLUTION1) Decide what must be maximized or minimized. We must maximize area. What we do not know are the rectangle’s dimensions, x and y.
xx
y
Blitzer, Intermediate Algebra, 5e – Slide #30 Section 8.3
Minimums & MaximumsCONTINUE
D 2) Express this quantity as a function in one variable. Because we must maximize area, we have A = xy. We need to transform this into a function in which A is represented by one variable. Because you have 200 feet of fencing, the sum of the lengths of the three sides of the rectangle that need to be fenced is 200 feet. This means that
2x + y = 200.We can solve this equation for y in terms of x, substitute the result into A = xy, and obtain A as a function in one variable. We begin by solving for y.
y = 200 – 2x Subtract 2x from both sides.
Blitzer, Intermediate Algebra, 5e – Slide #31 Section 8.3
Minimums & MaximumsCONTINUE
D Now we substitute 200 – 2x for y in A = xy.
A = xy = x(200 – 2x)
This function models the area, A (x), of any rectangle whose perimeter is 200 feet (and one side is not counted) in terms of one of its dimensions, x.
The rectangle and its dimensions are illustrated in the picture at the beginning of this exercise. Because A is now a function of x, we can write
A (x) = x(200 – 2x).
Blitzer, Intermediate Algebra, 5e – Slide #32 Section 8.3
Minimums & MaximumsCONTINUE
D 3) Write the function in the form . We apply the distributive property to obtain
cbxaxxf 2
.200222002200 22 xxxxxxxA
a = -2 b = 200
4) Calculate . If a < 0, the function has a maximum at
this value. The voice balloons show that a = -2 and b = 200.a
b2
50504
20022
2002
a
bx
Blitzer, Intermediate Algebra, 5e – Slide #33 Section 8.3
Minimums & MaximumsCONTINUE
D
5) Answer the question posed in the problem. We found that x = 50. The picture at the beginning of this exercise shows that the rectangle’s other dimension is 200 – 2x = 200 – 2(50) = 200 – 100 = 100 feet. The dimensions of the rectangle that maximize the enclosed area are 50 feet by 100 feet. The rectangle that gives the maximum area has an area of (50 feet) x (100 feet) = 5,000 square feet.
This means that the area, A (x), of a rectangle with a “3-sided” perimeter 200 feet is a maximum when the lengths of the two sides that are the same, x, are 50 feet.
In summary…
Blitzer, Intermediate Algebra, 5e – Slide #34 Section 8.3
We consider two standard forms for the quadratic function. In either form, it is easy to see whether the parabola opens upward or downward. We just consider the sign of a in either equation. In the first form that we considered, we could easily see the vertex of the parabola. In the second form, we could easily see the y intercept.You must decide which form is easiest for you to use in a given situation.
The vertex of the parabola is important. For a parabola opening upward, at the vertex we obtain a minimum function value. For a parabola opening downward, at the vertex we obtain a maximum function value.
Have you met Mini and Maxi? Mini is always smiling. Maxi is always frowning.
DONE
Blitzer, Intermediate Algebra, 5e – Slide #36 Section 8.3
Graphing Quadratic Functions
Graphs of Quadratic FunctionsThe graph of the quadratic function
is called a parabola. 0,2 acbxaxxf