9-7 Linear, Quadratic, and Exponential Models
Quick Review
Graphing data points or analyzing data numerically can help you find the best model. Linear data have a common first difference. Exponential data have a common ratio. Quadratic data have a common second difference.
Example
Graph the points (1, 4), (4, 2), (2, 3), (5, 3.5), and (6, 5). Which model is most appropriate?
A quadratic model is most appropriate.
Exercises
Graph each set of points. Which model is most appropriate for each data set?
-
(
−
3
,
0
)
,
(
1
,
4
)
,
(
−
1
,
6
)
,
(
2
,
0
)
open negative 3 comma 0 close comma open 1 comma 4 close comma open negative 1 comma 6 close comma open 2 comma 0 close
- (0, 6), (5, 2), (1, 4), (8, 1.5), (2, 3)
Write an equation to model the data.
-
x
|
y
|
−
1
negative 1
|
−
5
negative 5
|
0 |
−
2
negative 2
|
1 |
1 |
2 |
4 |
3 |
7 |
-
x
|
y
|
−
1
negative 1
|
2.5 |
0 |
5 |
1 |
10 |
2 |
20 |
3 |
40 |
9-8 Systems of Linear and Quadratic Equations
Quick Review
Systems of linear and quadratic equations can have two solutions, one solution, or no solution. These systems can be solved graphically or algebraically.
Example
What are the solutions of the system?
y
=
x
2
−
7
x
−
40
y
=
−
3
x
+
37
table with 2 rows and 1 column , row1 column 1 , y equals , x squared , minus 7 x minus 40 , row2 column 1 , y equals negative 3 x plus 37 , end table
y
=
x
2
−
7
x
−
40
−
(
y
=
−
3
x
+
37
)
0
=
x
2
−
4
x
−
77
Use elimination
.
Subtract the equations
.
0
=
(
x
−
11
)
(
x
+
7
)
Factor
.
x
−
11
=
0
or
x
+
7
=
0
x
=
11
or
x
=
−
7
Zero-Product Property
Solve for
x
.
table with 2 rows and 1 column , row1 column 1 , table with 1 row and 2 columns , row1 column 1 , table with 3 rows and 2 columns , row1 column 1 , y , column 2 equals , x squared , minus 7 x minus 40 , row2 column 1 , negative open y , column 2 equals negative 3 x plus 37 close , row3 column 1 , 0 , column 2 equals , x squared , minus 4 x minus 77 , end table , column 2 table with 3 rows and 1 column , row1 column 1 , cap useelimination . . , row2 column 1 , cap subtracttheequations . . , row3 column 1 , , end table , end table , row2 column 1 , table with 2 rows and 2 columns , row1 column 1 , 0 equals . open , x minus 11 , close . open , x plus 7 , close , column 2 cap factor , . , row2 column 1 , table with 2 rows and 5 columns , row1 column 1 , x minus 11 , column 2 equals 0 , column 3 or , column 4 x plus 7 , column 5 equals 0 , row2 column 1 , x , column 2 equals 11 , column 3 or , column 4 x , column 5 equals negative 7 , end table , column 2 table with 2 rows and 1 column , row1 column 1 , cap zerominuscap productcap property , row2 column 1 , cap solvefor . x . , end table , end table , end table
Find the corresponding y-values.
y
=
−
3
(
11
)
+
37
=
4
|
y
=
−
3
(
−
7
)
+
37
=
58
table with 1 row and 3 columns , row1 column 1 , y equals negative 3 , open 11 close , plus 37 equals 4 , column 2 vertical line , column 3 y equals negative 3 . open , negative 7 , close . plus 37 equals 58 , end table
The solutions are (11, 4) and
(
−
7
,
58
)
.
open negative 7 comma 58 close .
Exercises
Solve each system by graphing.
-
y
=
x
2
−
4
x
+
3
y
=
−
3
x
+
5
table with 2 rows and 1 column , row1 column 1 , y equals , x squared , minus 4 x plus 3 , row2 column 1 , y equals negative 3 x plus 5 , end table
-
y
=
x
2
−
2
x
−
1
y
=
−
x
−
1
table with 2 rows and 1 column , row1 column 1 , y equals , x squared , minus 2 x minus 1 , row2 column 1 , y equals negative x minus 1 , end table
-
y
=
−
2
x
2
+
x
+
2
y
=
x
table with 2 rows and 1 column , row1 column 1 , y equals negative 2 , x squared , plus x plus 2 , row2 column 1 , y equals x , end table
-
y
=
x
2
+
x
−
6
y
=
2
x
table with 2 rows and 1 column , row1 column 1 , y equals , x squared , plus x minus 6 , row2 column 1 , y equals 2 x , end table
Solve each system algebraically.
-
y
=
x
2
+
2
x
−
45
y
=
6
x
+
51
table with 2 rows and 1 column , row1 column 1 , y equals , x squared , plus 2 x minus 45 , row2 column 1 , y equals 6 x plus 51 , end table
-
y
=
x
2
−
12
x
+
33
y
=
4
x
−
30
table with 2 rows and 1 column , row1 column 1 , y equals , x squared , minus 12 x plus 33 , row2 column 1 , y equals 4 x minus 30 , end table
-
y
=
x
2
+
19
x
+
39
y
−
11
=
8
x
table with 2 rows and 1 column , row1 column 1 , y equals , x squared , plus 19 x plus 39 , row2 column 1 , y minus 11 equals 8 x , end table
-
y
=
x
2
+
5
x
−
40
y
+
1
=
−
5
x
table with 2 rows and 1 column , row1 column 1 , y equals , x squared , plus 5 x minus 40 , row2 column 1 , y plus 1 equals negative 5 x , end table
-
y
=
x
2
+
3
x
+
15
y
+
45
=
19
x
table with 2 rows and 1 column , row1 column 1 , y equals , x squared , plus 3 x plus 15 , row2 column 1 , y plus 45 equals 19 x , end table
-
y
=
x
2
+
11
x
+
51
y
=
−
10
x
−
57
table with 2 rows and 1 column , row1 column 1 , y equals , x squared , plus 11 x plus 51 , row2 column 1 , y equals negative 10 x minus 57 , end table
-
Writing Explain how you can use graphing to determine the number of solutions of a system of linear and quadratic equations.