Question

2
jack works after school. each day he earns a set amount, plus an hourl
(a)
hours\t1\t1.5\t2\t2.5\t3
pay\t18\t23\t28\t33\t38

part a
write a linear function f jack can use to determine his pay.
a ( f(x) = 10x + 8 )
b ( f(x) = 15x + 3 )
c ( f(x) = 0.5x + 1 )
d ( f(x) = x + 18 )

(b) part b
which is a reasonable domain?
a ( 0 leq x leq 6 )
b ( 0 leq x leq 24 )
c ( 0 leq x leq 68 )
d ( 0 leq x leq 248 )

Explanation:

Part A

Step1: Recall linear function form

A linear function is in the form \( f(x)=mx + b \), where \( m \) is the slope (hourly rate) and \( b \) is the y - intercept (set amount).

Step2: Calculate the slope \( m \)

Using two points from the table, e.g., \( (1,18) \) and \( (2,28) \). The slope formula is \( m=\frac{y_2 - y_1}{x_2 - x_1} \).
So \( m=\frac{28 - 18}{2 - 1}=\frac{10}{1}=10 \).

Step3: Find the y - intercept \( b \)

Using the point \( (1,18) \) and \( m = 10 \) in \( y=mx + b \).
\( 18=10\times1 + b \)
\( 18 - 10=b \)
\( b = 8 \).
So the linear function is \( f(x)=10x + 8 \), which is option A.

Part B

Step1: Understand the context

The domain represents the number of hours Jack can work. Since he works after school, a reasonable number of hours per day is limited.

Step2: Analyze the options

• Option B (\( 0\leq x\leq24 \)): 24 hours is a full day, which is not reasonable for after - school work.
• Option C (\( 0\leq x\leq68 \)): 68 hours is way too many for a daily after - school work.
• Option D (\( 0\leq x\leq248 \)): 248 hours is extremely unreasonable.
• Option A (\( 0\leq x\leq6 \)): 6 hours of after - school work per day is a reasonable amount.

Answer:

(Part A): A. \( f(x)=10x + 8 \)