Carnegie Learning Algebra II Student Skills Practice 1st Edition Chapter 2 Skills Practice
Page 294 Problem 1 Answer
The objective is to determine the equation for the quadratic function using the given graph.
We are given: The two x−intercepts (−7,0) ,(5,0) and one point (−4,−9)
Given: The two x−intercepts (−7,0),(5,0), and one point (−4,−9)
To obtain the values of a,b, and c, we will substitute the given points in the equation of the standard form, y=ax2+bx+c…(1).
Using the point(−7,0), we substitute x=−7 and y=0 into equation (1) and we will obtain an equation containing variables a,b,c,
we get,
0=a(−7)2+b(−7)+c
49a−7b+c=0…(2)
Using the point (5,0), we substitute x=5 and y=0 into equation (1) and we will obtain an equation containing variables a,b,c,we get,
0=a(5)2+b(5)+c
25a+5b+c=0…(3)
Using the point (−4,−9), we substitute x=−4 and y=−9 into equation (1) and we will obtain an equation containing variables a,b,c, we get,
−9=a(−4)2+b(−4)+c
16a−4b+c=−9…(4)
In order to obtain the values of the variables a,b,c, we will solve the obtained equations simultaneously.
Solving (2) and (3) simultaneously, we get,
49a− 7b+ c=0
25a+ 5b+ c=0
(-) (-) (-)
________________
24a−12b=0
b=2a
Solving (3) and (4),
9a+9b=9
b=1−a
Comparing both,
3a=1
a=1/3
Then, b=2/3.
And, c=7(2/3)−49(1/3)=14−49/3=−35/3
Therefore, a=1/3,b=2/3,c=−35/3.
Page 294 Problem 2 Answer
The objective is to determine the equation of the parabola in the vertex form using the given graph.
We are given: Vertex (−4,3) and y− intercept (0,11)
Given: Vertex (−4,3) and the y−intercept (0,11)
In order to obtain the distance from the axis of symmetry, we will substitute all the given values in the equation y=a(x−h)2+k, we get,
y=a(x+4)2+3
11=a(0+4)2+3
11=16a+3
16a=8
a=0.5
The equation in the vertex form is,
y=a(x−h)2+k
y=1/2(x+4)2+3
The equation is, y=1/2(x+4)2+3
Hence, the equation for the quadratic function using the given data is y=1/2(x+4)2+3.
Page 295 Problem 3 Answer
Given the points (2,0),(0,−12).
The aim is to find the equation.
The slope is, y=−12−0/0−2
=12/2
=6.
The equation is,
y+12=6(x)
y=6x−12
Therefore, the equation is y=6x−12.
Page 295 Problem 4 Answer
The objective is to determine the equation of the parabola in the vertex form using the given graph.
We are given: Vertex (−6,−1) and Point
Given: Vertex (−6,−1) and Point (−3,35)
To obtain the distance from the axis of symmetry, we will substitute all the given values in the equation y=a(x−h)2+k,
we get,
y=a(x+6)2−1
35=a(−3+6)2−1
35+1=9a
9a=36
a=4
The equation in the vertex form is,
y=a(x−h)2+k
y=4(x+6)2−1
The equation is, y=4(x+6)2−1
Hence, the equation for the quadratic function using the given data is y=4(x+6)2−1.
Page 296 Problem 5 Answer
The objective is to determine the equation for the quadratic function using the given data.
We are given: the three points which lie on the parabola are (5,−56),(1,−4),(−10,−26)
Given: Three points as (5,−56),(1,−4),(−10,−26)
We will obtain three sets of equations when we substitute the given points in the standard form of the parabola equations, we get,
For Point (5,−56):−56=25a+5b+c…(1)
For Point (1,−4):−4=a+b+c…(2)
For Point (−10,−26):−26=100a−10b+c…(3)
To evaluate the values of the missing variable, we will solve any two equations simultaneously
Adding (2) and (3), solving simultaneously, we get,
First, we will multiply (2) by 10
10a+10b+10c=−40
100a−10b+c=−26
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110a+11c=−66…(4)
___________________
Subtracting (1) and (2), solving simultaneously, we get,
First, we will multiply (2) by 5
25a+ 5b+ c= −56
5a+ 5b+ 5c= −20
(-) (-) (-) (+)
_________________
20a+4c=−36…(5)
_________________
Subtracting (4) and (5), solving simultaneously, we get,
First, we will multiply (4) by 4 and (5) by 11
440a+ 44c= −264
220a+ 44c= −396
(-) (-) (+)
_________________
220a=132⇒a=0.6
_________________
Now, we will substitute a=0.6 in (5), we get,
20(0.6)+4c=−36
4c=−36−12
c=−12
Now, on substituting a=0.6,c=−12 in (2),
we get,
0.6+b−12=−4
b=7.4
The equation in the standard form on substituting all the values is,
y=ax2+bx+c
y=0.6x2+7.4x−12
Hence, the equation for the quadratic function using the given data is y=0.6x2+7.4x−12.
Page 296 Problem 6 Answer
The objective is to determine the equation for the quadratic function using the given data.
We are given: the three points which lie on the parabola are (−8,8),(−4,6),(4,38)
Given: Three points as (−8,8),(−4,6),(4,38)
We will obtain three sets of equations when we substitute the given points in the standard form of the parabola equations, we get,
For Point (−8,8):8=64a−8b+c…(1)
For Point (−4,6):6=16a−4b+c…(2)
For Point (4,38):38=16a+4b+c…(3)
To evaluate the values of the missing variable, we will solve any two equations simultaneously
Adding (2) and (3), solving simultaneously, we get,
16a−4b+c=6
16a+4b+c=38
_______________
32a+2c=44…(4)
______________
Adding (1) and (3), solving simultaneously, we get,
First, we multiply (3) by 2
64a−8b+c=8
32a+8b+2c=76
_______________
96a+3c=84…(5)
________________
Subtracting (4) and (5), solving simultaneously, we get,
First, we will multiply (4) by 3 and (5) by 2
96a+ 6c= 132
192a+ 6c= 168
(-) (-) (-)
________________________
−96a=−36⇒a=0.375
________________________
Now, we will substitute a=0.375 in (4), we get,
32(0.375)+2c=44
2c=44−12
2c=32
c=16
Now, on substituting a=0.375,c=12 in (2), we get,
16(0.375)−4b+12=6
4b=12
b=3
The equation in the standard form on substituting all the values is,
y=ax2+bx+c
y=0.375x2+3x+12
Hence, the equation for the quadratic function using the given data is y=0.375x2+3x+12.
Page 296 Problem 7 Answer
The objective is to determine the equation for the quadratic function using the given data.
We are given: the three points which lie on the parabola are (−2,3),(2,−9),(5,−60)
Given: Three points as (−2,3),(2,−9),(5,−60)
We will obtain three sets of equations when we substitute the given points in the standard form of the parabola equations, we get,
For Point (−2,3):3=4a−2b+c…(1)
For Point (2,−9):−9=4a+2b+c…(2)
For Point (5,−60):−60=25a+5b+c…(3)
To evaluate the values of the missing variable, we will solve any two equations simultaneously
Adding (1) and (2), solving simultaneously, we get,
4a−2b+c=3
4a+2b+c=−9
______________
8a+2c=−6…(4)
______________
Subtracting (2) and (3), solving simultaneously, we get,
First, we will multiply (2) by 5 and (3) by 2
20a+ 10b+ 5c= 45
50a+ 10b+ 2c= −120
(-) (-) (-) (+)
___________________________
−30a+3c=165…(5)
____________________________
Subtracting (4) and (5), solving simultaneously, we get,
First, we multiply (4) by 3 and (5) by 2.
24a+6c=−18
−60a+6c=330
(+) (-) (-)
___________________________
84a=−348⇒a=−29/7
___________________________
Now substitute a=−29/7 in (4),
we get,8(−29/7)+2c=−6
2c=27.14285
c=13.57142
Now, substitute a=4.142857,c=13.57142 in (2),
we get,
4(4.142857)+2b+13.57142=−9
b=13.571424
The equation in the standard form on substituting all the values is, y=ax2+bx+c
y=4.142857x2+13.571424x+13.57142
Hence, the equation for the quadratic function using the given data is y=4.142857x2+13.571424x+13.57142.
Page 298 Problem 8 Answer
Given: (−2,−2),(1,−5),(2,−18)
To find: The system of equations
A quadratic equation through given points.
Put these points in f(x)=ax2+bx+c
We will get three equations in a,b,c. Solve it by the substitution method.
For (−2,−2): 4a−2b+c=−2…(1)
For (1,−5): a+b+c=−5…(2)
For (2,−18): 4a+2b+c=−18…(3)
a=−2+2b−c/4 From 1…(5)
6b+3c−2/4=−5 From 2…(4)
4⋅−2+2b−c/4+2b+c=−18 From 3
4b−2=−18
b=−4
Putting in(4),3c−26/4=−5
c=2
a=−2+2(−4)−2/4 From 5
a=−3
⇒a=−3,c=2,b=−4
The required quadratic equation: −3x2−4x+2
The quadratic equation: −3x2−4x+2
Page 300 Problem 9 Answer
Given: (−1,2),(4,27),(−3,20)
To find: The system of equations
A quadratic equation through given points.
Put these points in f(x)=ax2+bx+c
We will get three equations in a,b,c. Solve it by the substitution method.
For (−1,2): a−b+c=2…(1)
For (4,27): 16a+4b+c=27…(2)
For (−3,20): 9a−3b+c=20…(3)
Substitutea=2+b−c…(4)
20b−15c+32=27 From 2&4
b=3c−1/4…(5)
6b−8c+18=20 From 3&4
6⋅3c−1/4−8c+18=20 From 5
−7c−3/2+18=20
c=−1
b=3(−1)−1/4 From 5
a=2−1−(−1) From 4
a=2
⇒a=2,c=−1,b=−1
The required quadratic equation: 2x2−x−1
The quadratic equation: 2x2−x−1
Page 301 Problem 10 Answer
Given: (5,−6),(−2,8),(3,4)
To find: The system of equations
A quadratic equation through given points.
Put these points in f(x)=ax2+bx+c
We will get three equations in a,b,c. Solve it by the substitution method.
For (5,−6): 25a+5b+c=−6…(1)
For (−2,8):
4a−2b+c=8…(2)
For (3,4):
9a+3b+c=4…(3)
Substitutea=−6−5b−c/25 From 1…(4)
4⋅−6−5b−c/25−2b+c=8 From 2&4
−70b+21c−24/25=8
b=−−3c+32/10…(5)
30b+16c−54/25=4 From 3&4
30(−−3c+32/10)+16c−54/25
=4 From 5
c−6=4
c=10
b=−−3⋅10+32/10 From 5
b=−1/5
a=−6−5(−1/5)−10/25 From 4
a=−3/5
⇒a=−3/5,c=10,b=−1/5
The required quadratic equation: −3/5x2−1/5x+10
The quadratic equation: −3/5x2−1/5x+10
Page 302 Problem 11 Answer
Given: (1,17),(−1,−9),(2,105)
To find: The system of equations
A quadratic equation through given points.
Put these points in f(x)=ax2+bx+c
We will get three equations in a,b,c. Solve it by the substitution method.
For (1,17): a+b+c=17…(1)
For (−1,−9): a−b+c=−9…(2)
For (2,105): 4a+2b+c=105…(3)
a=17−b−c For 1…(4)
17−b−c−b+c=−9 From 2&4
−2b+17=−9
b=13
4(17−b−c)+2b+c=105
−2b−3c+68=105
−2⋅13−3c+68=105 Substituteb=13
−3c+42=105
c=−21
a=17−13−(−21) From 4
a=25
⇒a=25,c=−21,b=13
The required quadratic equation: 25x2+13x−21
The quadratic equation: 25x2+13x−21