Linear Functions Unit Test
Algebra 1 • Unit 4 Assessment
Linear Functions & Equations
Name:
Date:
Period:
Score: / 100
Directions: Read each question carefully. Select the best answer or complete the designated blanks. Show all necessary work. Page 1 of 5
1
A graph of a linear function is shown on the coordinate grid below.
x y 2 4 6 -2 -4 2 4 6 -2 -4
Write an equation in slope-intercept form that best represents this graph. Fill in the blanks below.
y = x +
2
The Westview Robotics Club hosted a regional showcase to raise funds for competition gear. They spent $180 on printed banners and $270 on safety equipment and pit decorations. Each spectator ticket to the showcase was sold for $8. Which linear function represents the total profit, y, the robotics club makes if they sell x tickets?
A. y = 8x + 450
B. y = 450x − 8
C. y = 8x − 450
D. y = 8x − 180
3
Which equation describes the line through the points (3, −2) and (5, 4) in point-slope form?
A. y − 4 = −⅓(x − 5)
B. y − 3 = −3(x + 2)
C. y − 4 = ⅓(x − 5)
D. y + 2 = 3(x − 3)
Algebra 1 • Unit 4: Linear Functions Assessment Turn page for Questions 4–6
Unit 4: Linear Functions Exam Page 2 of 5
4
The first five terms of an arithmetic sequence are shown below:
a1 = 28 a2 = 19 a3 = 10 a4 = 1 a5 = −8
Select the correct numbers from the bank below to create an equation that can be used to find the _n_th term of the sequence, an. Not all numbers will be used.
−9 9 −28 28 37 −37
an =
n +
5
What is the equation in slope-intercept form of the line that crosses the x-axis at 4 and is perpendicular to the line represented by \( y = -\frac{2}{3}x + 5 \)?
A. \( y = -\frac{2}{3}x + 4 \)
B. \( y = \frac{3}{2}x - 6 \)
C. \( y = -\frac{2}{3}x - 6 \)
D. \( y = \frac{3}{2}x + 4 \)
6
The graph of a horizontal line is shown on the coordinate plane below.
What is the equation and the slope of the line? Select the correct values from the tile options. Not all options will be used.
x y 4 -4 4 -4
x = −2 y = −2 y = −2x undefined 0 −2
Slope:
Equation:
Algebra 1 • Unit 4: Linear Functions Assessment Turn page for Questions 7–10
Unit 4: Linear Functions Exam Page 3 of 5
7
Which equation in point-slope form best describes the line that is parallel to \( y = \frac{5}{2}x + 8 \) and passes through the point \( (2, -7) \)?
A. \( y + 7 = \frac{5}{2}(x - 2) \)
B. \( y + 7 = -\frac{5}{2}(x - 2) \)
C. \( y + 7 = \frac{2}{5}(x - 2) \)
D. \( y + 7 = -\frac{2}{5}(x - 2) \)
8
Elena works as an after-school academic tutor. Her daily wage, w, is based on the number of hours she tutors, h, at her standard rate of $18.50 per hour, plus a fixed $15 daily travel and materials stipend. Which equation best describes the relationship between her daily wage, w, and hours worked, h?
A. w = 15h + 18.50
B. w = 18.50h + 15
C. w = 18.50h + 15h
D. w = 18.50h
9
The graph models the linear relationship between the number of monthly payments made on a car loan and the remaining balance in dollars left to pay on the loan.
Car Loan Balance
30,000 25,000 20,000 15,000 10,000 5,000 0 10 20 30 40 50 60 Number of Monthly Payments Remaining Balance ($) (10, 20,000) (25, 14,000)
Write an equation that best represents this situation.
y = x +
10
The table represents some points on the graph of a linear function. Which function represents this relationship?
A. \( y = \frac{4}{3}x + 2 \)
B. \( y = \frac{4}{3}x - 5 \)
C. \( y = \frac{3}{4}x + 7 \)
D. \( y = \frac{3}{4}x - 3 \)
Algebra 1 • Unit 4: Linear Functions Assessment Turn page for Questions 11–13
Unit 4: Linear Functions Exam Page 4 of 5
11
A professional photography studio charges a flat studio setup fee plus a constant hourly shooting rate for portrait sessions. One client booked the studio for 2 hours for a total of $190. Another client booked the studio for 6 hours for a total cost of $450. Which linear equation represents the total cost in dollars, y, for booking the studio for x hours?
A. y = 15x + 320
B. y = 75x
C. y = 65x + 190
D. y = 65x + 60
12
Which equation is equivalent to \( 5x - y = 13 \) when solved for y?
A. y = −5x + 13
B. y = −5x − 13
C. y = 5x + 13
D. y = 5x − 13
13
Graph the following linear function: \( -3y = -2x + 6 \). Show your algebraic work to solve for slope-intercept form, then correctly plot at least TWO points on the coordinate plane and connect them with a straight line.
Show algebraic steps:
x y 2 4 -2 -4 2 4 -2 -4
Algebra 1 • Unit 4: Linear Functions Assessment Turn page for Questions 14–16
Unit 4: Linear Functions Exam Page 5 of 5
14
The values in the table represent a linear relationship between x and y. What is the rate of change of y with respect to x?
A. 17
B. −6
C. −3
D. 21
Use the graph below to answer Questions 15 and 16. The graph displays line m and point K.
x y 6 -6 6 -6 m K
15
Which equation best represents the slope-intercept form of a line that passes through point K and is parallel to line m?
A. \( y = \frac{1}{3}x + 1 \)
B. \( y = -\frac{1}{3}x - 2 \)
C. \( y = -\frac{1}{3}x + 3 \)
D. \( y = \frac{1}{3}x + 3 \)
16
Constructed Response: Explain how you determined the equation of the line that passes through point K and is parallel to line m. Complete your explanation using the sentence stem below.
Sentence Stem: I determined the equation of the new line by...
Algebra 1 • Unit 4: Linear Functions Assessment End of Assessment • Verify all responses
Linear Functions Test Key
Teacher Guide • Solutions & Rubrics
Linear Functions Test Key
MASTER KEY
Total Points: 100
Grading Overview: Questions 1–14 are 6 pts each (84 pts). Question 15 is 6 pts. Question 16 is 10 pts. Page 1 of 4
1 Slope-Intercept from Graph (Positive Slope)
6 Points
Answer: \( y = 2x + 3 \) (First blank: 2, Second blank: 3)
Worked Solution: Identify the y-intercept from the graph at \( (0, 3) \), so \( b = 3 \). Locate a second lattice point such as \( (2, 7) \) (or \( (1, 5) \)). Calculate slope: \( m = \frac{7 - 3}{2 - 0} = \frac{4}{2} = 2 \) (positive slope). Combining yields \( y = 2x + 3 \).
2 Linear Profit Function
6 Points
Answer: C • \( y = 8x - 450 \)
Worked Solution: Total expenses = \( \$180 + \$270 = \$450 \) (fixed cost subtracted from revenue). Revenue per ticket = \( \$8 \). Profit = Revenue − Expenses = \( 8x - 450 \). Distractor A adds expenses; Distractor B reverses rate and expenses.
3 Point-Slope Form through Two Points
6 Points
Answer: D • \( y + 2 = 3(x - 3) \)
Worked Solution: Slope \( m = \frac{4 - (-2)}{5 - 3} = \frac{6}{2} = 3 \). Using point \( (3, -2) \) in point-slope formula \( y - y_1 = m(x - x_1) \) gives \( y - (-2) = 3(x - 3) \implies y + 2 = 3(x - 3) \).
4 Arithmetic Sequence Explicit Rule
6 Points
Answer: \( a_n = \mathbf{-9}n + \mathbf{37} \) (First box: −9, Second box: 37)
Worked Solution: Common difference \( d = 19 - 28 = -9 \). Use formula \( a_n = a_1 + (n - 1)d = 28 + (n - 1)(-9) = 28 - 9n + 9 = -9n + 37 \). Alternatively, zeroth term \( a_0 = 28 - (-9) = 37 \).
5 Perpendicular Line Equation
6 Points
Answer: B • \( y = \frac{3}{2}x - 6 \)
Worked Solution: The given line has slope \( m_1 = -\frac{2}{3} \). Perpendicular slope is opposite reciprocal: \( m_2 = \frac{3}{2} \). The line crosses the x-axis at 4, giving point \( (4, 0) \). Substitute: \( 0 = \frac{3}{2}(4) + b \implies 0 = 6 + b \implies b = -6 \). Equation is \( y = \frac{3}{2}x - 6 \).
Algebra 1 • Unit 4 Assessment Answer Key Turn page for Questions 6–10
Unit 4 Answer Key • Solutions Page 2 of 4
6 Horizontal Line Properties (y = −2)
6 Points (3 pts each)
Slope: 0 | Equation: y = −2
Worked Solution: Horizontal lines have zero rate of vertical change (\( \Delta y = 0 \)), so slope is \( 0 \) (HOY: Horizontal, 0 slope, y = constant). Every point on this horizontal line has a y-coordinate of −2, yielding equation \( y = -2 \).
7 Parallel Line in Point-Slope Form
6 Points
Answer: A • \( y + 7 = \frac{5}{2}(x - 2) \)
Worked Solution: Parallel lines share the same slope, so \( m = \frac{5}{2} \). Substituting the given point \( (x_1, y_1) = (2, -7) \) into point-slope form: \( y - (-7) = \frac{5}{2}(x - 2) \implies y + 7 = \frac{5}{2}(x - 2) \).
8 Linear Wage Relationship
6 Points
Answer: B • \( w = 18.50h + 15 \)
Worked Solution: The tutoring rate of $18.50 per hour is the constant rate of change (slope \( m = 18.50 \)), multiplied by hours \( h \). The fixed travel stipend of $15 is the constant start value (y-intercept \( b = 15 \)).
9 Equation from Context Graph
6 Points
Answer: \( y = -400x + 24000 \) (First blank: −400, Second blank: 24,000 or 24000)
Worked Solution: Using points \( (10, 20000) \) and \( (25, 14000) \): \[ m = \frac{14000 - 20000}{25 - 10} = \frac{-6000}{15} = -400 \text{ dollars/month} \] Find initial balance \( b \): \( 20000 = -400(10) + b \implies 20000 = -4000 + b \implies b = 24000 \). Equation: \( y = -400x + 24000 \).
10 Function Rule from Data Table
6 Points
Answer: B • \( y = \frac{4}{3}x - 5 \)
Worked Solution: Calculate slope using first two rows: \( m = \frac{7 - (-13)}{9 - (-6)} = \frac{20}{15} = \frac{4}{3} \). Confirm with third row: \( \frac{27 - 7}{24 - 9} = \frac{20}{15} = \frac{4}{3} \). Solve for b: \( 7 = \frac{4}{3}(9) + b \implies 7 = 12 + b \implies b = -5 \). Equation: \( y = \frac{4}{3}x - 5 \).
Algebra 1 • Unit 4 Assessment Answer Key Turn page for Questions 11–14
Unit 4 Answer Key • Solutions Page 3 of 4
11 Linear Model from Two Data Points
6 Points
Answer: D • \( y = 65x + 60 \)
Worked Solution: Points are \( (2, 190) \) and \( (6, 450) \). Hourly rate \( m = \frac{450 - 190}{6 - 2} = \frac{260}{4} = \$65/\text{hr} \). Setup fee \( b = 190 - 65(2) = 190 - 130 = \$60 \). Equation: \( y = 65x + 60 \).
12 Transforming to Slope-Intercept Form
6 Points
Answer: D • \( y = 5x - 13 \)
Worked Solution: Given \( 5x - y = 13 \). Subtract \( 5x \) from both sides: \( -y = -5x + 13 \). Multiply or divide through by \( -1 \): \( y = 5x - 13 \).
13 Graphing a Linear Function
6 Points
Answer: Equation \( y = \frac{2}{3}x - 2 \) • Correct points and line plotted
Worked Solution: Divide all terms in \( -3y = -2x + 6 \) by \( -3 \): \[ \frac{-3y}{-3} = \frac{-2x}{-3} + \frac{6}{-3} \implies y = \frac{2}{3}x - 2 \]
Scoring Breakdown (6 pts):
- 2 pts: Correct algebraic isolation of \( y = \frac{2}{3}x - 2 \).
- 2 pts: At least two correct points accurately plotted, such as \( (0, -2) \), \( (3, 0) \), or \( (6, 2) \).
- 2 pts: Straight continuous line drawn extending through both plotted points with arrows.
14 Rate of Change with Decimals
6 Points
Answer: C • −3
Worked Solution: Pick any two table pairs, e.g., \( (-4.5, 32) \) and \( (0.5, 17) \): \[ \text{Rate of Change} = \frac{\Delta y}{\Delta x} = \frac{17 - 32}{0.5 - (-4.5)} = \frac{-15}{5.0} = -3 \] Verification on next interval: \( \frac{8 - 17}{3.5 - 0.5} = \frac{-9}{3} = -3 \).
Algebra 1 • Unit 4 Assessment Answer Key Turn page for Questions 15–16 & Rubric
Unit 4 Answer Key • Solutions & Rubric Page 4 of 4
15 Parallel Line from Visual Graph
6 Points
Answer: C • \( y = -\frac{1}{3}x + 3 \)
Worked Solution: Line m passes through \( (0, -2) \) and \( (3, -3) \), giving slope \( m = \frac{-3 - (-2)}{3 - 0} = -\frac{1}{3} \). Because the line is parallel, it also has slope \( m = -\frac{1}{3} \). Point K is at \( (3, 2) \). Substitute into point-slope form: \[ y - 2 = -\frac{1}{3}(x - 3) \implies y - 2 = -\frac{1}{3}x + 1 \implies y = -\frac{1}{3}x + 3 \]
16 Constructed Response Explanation
10 Points Total
Exemplar Student Response:
“I determined the equation of the new line by first finding the slope of line m using points (0, −2) and (3, −3), which gave m = −1/3. Because parallel lines have equal slopes, my new line also has a slope of −1/3. Next, I identified point K at (3, 2) and substituted the point and slope into point-slope form: y − 2 = −1/3(x − 3). Finally, I solved for y in slope-intercept form by distributing and adding 2 to both sides, which resulted in y = −1/3x + 3.”
| Score |
Criteria |
| 10 pts |
Identifies slope \( -\frac{1}{3} \), explains parallel lines have identical slopes, identifies point \( K(3, 2) \), and explains algebraic steps to solve for intercept \( 3 \). |
| 7–8 pts |
Correctly states slope and parallel relationship, but contains minor algebraic imprecision or omits how the constant \( 3 \) was calculated. |
| 4–6 pts |
Mentions parallel lines have the same slope or identifies point \( K \), but fails to provide a coherent step-by-step algebraic explanation. |
| 1–3 pts |
Minimal attempt; relies solely on restating the sentence stem without valid mathematical justification. |
TEKS & CCSS Standards Alignment:
• Slope & Representations: A.2A, A.2B, A.2C (CCSS: HSF-IF.B.4, HSF-IF.B.6, HSF-LE.A.2)
• Parallel & Perpendicular: A.2E, A.2F, A.2G (CCSS: HSG-GPE.B.5)
• Arithmetic Sequences: A.12D (CCSS: HSF-BF.A.2)
• Modeling & Profit: A.2H, A.2I (CCSS: HSA-CED.A.2)
Algebra 1 • Unit 4 Assessment Answer Key End of Answer Key
