EVALUATE EACH FUNCTIONS FOR THE GIVEN INPUT VALUES

Find the value of y given rule :

Example 1  :

y = 8x+5

when (i) x  =  3 (ii) x  =  7 (iii) x  =  10 1/2

Solution :

Given rule, y = 8x+5

(i) when x = 3

By applying x = 3 in given rule, we get

y = 8x + 5

y = 8(3) + 5

y = 24+5

y = 29

Therefore, the value of y is 29

(ii) x = 7

By applying x = 7 in given rule, we get

y = 8x + 5

y = 8(7) + 5

y = 56 + 5

y = 61

Therefore, the value of y is 61

(iii) x  =  10 1/2

Since the value of x is mixed fraction, we convert it into improper. 

10 1/2  =  13/2

By applying x  =  13/2 in given rule, we get

y  =  8x+5

y  =  8(13/2)+5

y  =  52+5

y  =  57

Therefore, the value of y is 57

Example 2 :

y  =  21-4x

when (i) x  =  0 (ii) x  =  2 1/2 (iii) x  =  7

Solution :

Given rule, y  =  21-4x

(i) x  =  0

By applying x  =  0 in given rule, we get

y  =  21-4x

y  =  21-4(0)

y  =  21

Therefore, the value of y is 21

(ii) x  =  2 1/2

Converting the mixed fraction into improper fraction,

2 1/2  =  5/2

By applying x  =  5/2 in given rule, we get

y  =  21-4x

y  =  21-4(5/2)

y  =  21-10

y  =  11

Therefore, the value of y is 11

(iii) x  =  7

By applying x  =  7 in given rule, we get

y  =  21-4x

y  =  21-4(7)

y  =  21-28

y  =  -7

Therefore, the value of y is -7

Example 3 :

y  =  (3x+4)/2

when (i) x  =  2 (ii) x  =  6 (iii) x  =  11

Solution :

Given rule, y  =  (3x+4)/2

(i) x  =  2

By applying x  =  2 in given rule, we get

y  =  (3x+4)/2

y  =  [3(2)+4]/2

y  =  (6+4)/2

y  =  (10/2)

y  =  5

Therefore, the value of y is 5

(ii) x  =  6

By applying x  =  6 in given rule, we get

y  =  (3x+4)/2

y  =  [3(6)+4]/2

y  =  (18+4)/2

y  =  (22/2)

y  =  11

Therefore, the value of y is 11

(iii) x  =  11

By applying x  =  11 in given rule, we get

y  =  (3x+4)/2

y  =  [3(11)+4]/2

y  =  (33+4)/2

y  =  (37/2)

y  =  18.5

Therefore, the value of y is 18.5

Example 4 :

y  =  2(x+3)-1

when (i) x  =  4 (ii) x  =  0 (iii) x  =  6 1/2

Solution :

Given rule, y  =  2(x+3)-1

(i) x  =  4

By applying x  =  4 in given rule, we get

y  =  2(x+3)-1

y  =  2(4+3)-1

y  =  2(7)-1

y  =  14-1

y  =  13

Therefore, the value of y is 13

(ii) x  =  0

By applying x  =  0 in given rule, we get

y  =  2(x+3)-1

y  =  2(0+3)-1

y  =  2(3)-1

y  =  6-1

y  =  5

Therefore, the value of y is 5

(iii) x  =  6 1/2

Changing into mixed fraction, we get

6 1/2  =  13/2

By applying x  =  13/2 in given rule, we get

y  =  2(x+3)-1

y  =  2(13/2+3)-1

y  =  2(19/2)-1

y  =  19-1

y  =  18

Therefore, the value of y is 18.

Example 5 :

s = 40 + 3t

The equation gives the speed s, in miles per hour, of a certain car t seconds after it began to accelerate. What is the speed, in miles per hour, of the car 5 seconds after it began to accelerate?

A) 40    B) 43    C) 45    D) 55

Solution :

s - speed, t - number of seconds

When t = 5 (5 seconds after)

s = 40 + 3t

s = 40 + 3(5)

= 40 + 15

= 55

So, option D is correct.

Example 6 :

The function p is defined by p = 7n³ . What is the value of n when p (n) is equal to 56?

A) 2     B) 8/3    C) 7    D) 8

Solution :

p = 7n³

When p(n) = 56

56 = 7n³

n³ = 56/7

n³ = 8

n = 2

So, option A is correct.

Example 7 :

Vivian bought party hats and cupcakes for $71. Each package of party hats cost $3, and each cupcake cost $1. If Vivian bought 10 packages of party hats, how many cupcakes did she buy?

Solution :

Cost of party hats and cupcakes = $71

Cost of each hat = $3

Cost of each cupcake =$1

Let x be the hat and y be the cupcake,

3x + 1y = 71 -----(1)

Number of party packs = 10

3(10) + y = 71

y = 71 - 30

y = 41

Example 8 :

y = -1.5

y = x² + 8 x + a

In the given system of equations, a is a positive constant. The system has exactly one distinct real solution. What is the value of a?

Solution :

y = -1.5 ----(1)

y = x² + 8 x + a ----(2)

Applying the value of y, we get

-1.5 = x² + 8 x + a

x² + 8 x + a + 1.5 = 0

Since it has real solution, b² - 4ac ≥ 0

a = 1, b = 8 and c = a + 1.5

8² - 4(1)(a + 1.5) ≥ 0

64 - 4a - 6 = 0

2a - 6 = 0

2a = 6

a = 6/2

a = 3

Example 9 :

The function his defined by h(x) = 4x + 28. The graph of y = h(x) in the -plane has an x-intercept at (a, 0) and a y-intercept at (0, b), where a and b are constants. What is the value of a + b?

A) 21      B)  28       C) 32       D) 35

Solution :

h(x) = 4x + 28

x-intercept is at (a, 0)

0 = 4a + 28

4a = -28

a = -28/4

a = -7

y-intercept is at (0, b).

b = 4(0) + 28

b = 28

a + b = -7 + 28

a + b = 21

So, option A is correct.

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