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Quizzes > Mathematics > Algebra

Take the Algebra 2 Section 1.3 & 1.4 Quiz Now!

Ready to Nail Your Algebra 2 1.3 Answers? Dive In Now!

Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art illustration of Algebra 2 Section 1.3 and 1.4 quiz on coral background with lined spaces for answers

This Algebra 2 Lesson 1.3 Practice B quiz helps you work with lines: graph them, find slope-intercept form, and use point-slope. Practice to spot gaps before a test or homework check, and build speed and accuracy. When you finish, try more in Chapter 2 practice or browse extra Algebra 2 questions.

What is the slope of the line 2x - 3y = 6?
3/2
-3/2
2/3
-2/3
To find the slope, rewrite the equation in y=mx+b form: 2x - 3y = 6 becomes 3y = 2x - 6, so y = (2/3)x - 2. The coefficient of x is the slope, which is 2/3. This method applies to any linear equation when solving for y. .
What are the coordinates of the y-intercept of the line y = -4x + 7?
(7, 0)
(0, 7)
(0, -7)
(-4, 0)
The y-intercept occurs when x=0, so substituting gives y=7, yielding the point (0, 7). Other options confuse x- and y-intercepts or sign of y. Identifying the intercept requires setting x to zero in the slope-intercept form. .
Evaluate f(x) = 3x + 2 for x = 5.
-7
13
11
17
Substitute x=5 into the function: f(5)=3(5)+2=15+2=17. This direct substitution is standard for evaluating linear functions. It confirms that you handle addition and multiplication in the correct order. Function evaluation guide.
Is the relation { (1,2), (2,3), (3,4), (2,5) } a function?
True
False
A relation is a function if each input has exactly one output. Here, the input 2 maps to both 3 and 5, violating the definition. Therefore, the relation is not a function. Understanding functions vs. relations.
Write the equation of the line in slope-intercept form that passes through (1, 2) and has a slope of 3.
y = x + 2
y = 2x + 3
y = 3x - 1
y = 3x + 1
Use point-slope form: y - 2 = 3(x - 1). Distribute and simplify: y - 2 = 3x - 3, so y = 3x - 1. This matches the slope and given point. .
Find the equation of the line perpendicular to y = (1/2)x + 4 and passing through (4, -1).
y = -2x + 7
y = 2x - 9
y = 1/2x - 3
y = -1/2x + 1
The slope of the given line is 1/2, so the perpendicular slope is -2. Use point-slope: y + 1 = -2(x - 4), giving y = -2x + 8 - 1 ? y = -2x + 7. .
What is the domain of the function f(x) = 1/(x - 3)?
[3, ?)
(-?, 3) U (3, ?)
(-?, 3]
(-?, ?)
The denominator cannot be zero, so x ? 3. All other real numbers are allowed. Thus the domain excludes 3, which is written as (-?, 3) ? (3, ?). .
Given f(x) = x² - 1 and g(x) = 2x + 3, what is f(g(1))?
-2
8
10
24
First compute g(1)=2(1)+3=5, then f(5)=5²-1=25-1=24. This shows you must apply the inner function before the outer one. Function composition.
Find the inverse function of f(x) = 2x - 5.
(x - 5)/2
(5 - x)/2
(x + 5)/2
2x + 5
Swap x and y: x = 2y - 5, then solve for y: 2y = x + 5, so y = (x + 5)/2. Thus f?¹(x)= (x+5)/2. Inverse functions method.
Solve the system of equations: y = 2x + 1 and y = -x + 4.
(-1, -3)
(1, 3)
(1, -3)
(3, 1)
Set 2x + 1 = -x + 4, giving 3x = 3 and x = 1. Substitute back to get y = 2(1) + 1 = 3. Thus the solution is (1,3). .
Solve the equation |2x - 5| = 7.
x = -1
x = 6
x = 6 or x = -1
x = ±3.5
Set 2x - 5 = 7 ? 2x = 12 ? x = 6, and 2x - 5 = -7 ? 2x = -2 ? x = -1. Absolute-value equations yield two cases. .
Find the inverse of the function f(x) = (2x + 1)/(x - 3).
(3x + 1)/(x - 2)
(1 - 2x)/(x + 3)
(x + 1)/(2 - x)
(3x - 1)/(x + 2)
Start with y = (2x+1)/(x-3), swap x and y, then solve: x(y-3) = 2y + 1 ? xy - 3x = 2y + 1 ? xy - 2y = 3x + 1 ? y(x - 2) = 3x + 1 ? y = (3x + 1)/(x - 2). Thus f?¹(x) = (3x+1)/(x-2). Advanced inverse functions.
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Study Outcomes

  1. Understand slope-intercept relationships -

    By reviewing the lesson 1.3 practice B Algebra 2 answers, you will grasp how the slope and y-intercept define and graph linear equations.

  2. Solve one-step and multi-step equations -

    You will sharpen your skills in solving linear equations efficiently, ensuring accuracy on algebra 2 chapter 1 and 2 tests.

  3. Distinguish parallel and perpendicular lines -

    Using algebra 2 1.3 answers, you will learn to identify and write equations for lines with given slopes or relationships.

  4. Apply direct variation concepts -

    You will deepen your understanding of direct variation by recognizing proportional relationships and writing corresponding equations.

  5. Interpret quiz results -

    Through the free Algebra II chapter 1 test style questions, you will analyze your errors and reinforce key concepts in linear functions.

  6. Evaluate problem-solving strategies -

    You will compare different solution methods to find the most efficient approach for solving linear equations and line-related problems.

Cheat Sheet

  1. Slope-Intercept Refresher -

    Mastering y = mx + b is essential for lesson 1.3 practice B Algebra 2 answers; m represents the slope and b is the y-intercept. For example, in y = 3x - 2, the line rises 3 units for every 1 unit it runs to the right, crossing the y-axis at - 2. This structure underpins most linear problems in your algebra 2 chapter 1 and 2 test.

  2. Identifying Parallel Lines -

    Parallel lines share the same slope but different intercepts, so if two equations both have m = 5, they'll never meet. A quick check on the algebra ii chapter 1 test: compare slopes directly rather than intercepts. Remember the mnemonic "Same Slope, Different Hope" to lock in the concept!

  3. Finding Perpendicular Lines -

    Perpendicular slopes are negative reciprocals: if one line has m = 2/3, its perpendicular partner has m = - 3/2. To practice, rewrite given lessons from Khan Academy or MIT OpenCourseWare to solidify "flip and change the sign." This trick will boost your confidence when tackling the algebra 2 chapter test.

  4. Direct Variation Basics -

    Direct variations follow y = kx, where k is the constant of variation; there's no intercept since b = 0. If y varies directly as x and k = 4, then y = 4x gives an immediate ratio of 4:1. Use this in lesson 1.3 practice b algebra 2 answers to identify proportional relationships swiftly.

  5. Interpreting Solutions Skillfully -

    Always check for special cases: parallel lines yield no solution, coincident lines yield infinite solutions, and unique slopes yield one solution. During your algebra 2 1.3 answers review, graph each equation pair or use substitution to confirm the intersection behavior. This layered approach ensures accuracy on the algebra 2 chapter 1 and 2 test.

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