Carnegie Learning Algebra II – 1st Edition Chapter 2 Exercise 2.5 Solutions

Carnegie Learning Algebra II – 1st Edition Chapter 2 Exercise 2.5 Solutions

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.

Carnegie Learning Algebra Ii Chapter 2 Exercise 2.5 Solutions

Carnegie Learning Algebra II – 1st Edition Chapter 2 Exercise 2.5 Solutions 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.

Carnegie Learning Algebra II – 1st Edition Chapter 2 Exercise 2.5 Solutions 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.

Carnegie Learning Algebra II – 1st Edition Chapter 2 Exercise 2.5 Solutions 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.

Exercise 2.5 Chapter 2 Carnegie Learning Answers

Carnegie Learning Algebra II – 1st Edition Chapter 2 Exercise 2.5 Solutions 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
___________________
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.

Carnegie Learning Algebra II – 1st Edition Chapter 2 Exercise 2.5 Solutions 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.

Carnegie Learning Algebra II – 1st Edition Chapter 2 Exercise 2.5 Solutions 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.

Algebra Ii Chapter 2 Exercise 2.5 Detailed Solutions

Carnegie Learning Algebra II – 1st Edition Chapter 2 Exercise 2.5 Solutions 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

Carnegie Learning Algebra II – 1st Edition Chapter 2 Exercise 2.5 Solutions 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)

Substitute​a=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

Carnegie Learning Chapter 2 Exercise 2.5 Explained

Carnegie Learning Algebra II – 1st Edition Chapter 2 Exercise 2.5 Solutions 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)

Substitute​a=−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

Carnegie Learning Algebra II – 1st Edition Chapter 2 Exercise 2.5 Solutions 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 Substitute​b=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