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Quizzes > Mathematics

Absolute Value Function Quiz: Math 144 Quiz 1

Quick, free absolute value quiz to test your skills. Instant results.

Editorial: Review CompletedCreated By: Tarek Al-AhejriUpdated Aug 28, 2025
Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art illustration for Math 144 Quiz 1 on absolute functions on a dark blue background

Use this Math 144 Quiz 1 to practice absolute value functions, from equations and inequalities to graph reading. Get instant feedback and pinpoint what to review next. For extra help, try absolute value practice, review writing absolute value as piecewise, or check absolute value inequality properties.

What are the solutions to the equation |x| = 5?
x = -5
x = 5
x = 5 or x = -5
No solution
The equation |x| = 5 means x is 5 units away from zero on the number line, giving x = 5 or x = -5. Absolute value measures distance and always yields a non-negative result. No other values satisfy the distance requirement.
Solve the equation |x - 3| = 7.
x = 4 or x = -10
x = -3 or x = -7
x = 3 or x = 7
x = 10 or x = -4
For |x - 3| = 7, set x - 3 = 7 or x - 3 = -7, yielding x = 10 or x = -4. This arises from the definition of absolute value as distance from zero. The other options correspond to incorrect manipulations.
Find the solutions for |2x + 4| = 8.
x = 2 or x = 6
x = -2 or x = -6
x = -2 or x = 6
x = 2 or x = -6
Two cases arise: 2x + 4 = 8 gives x = 2; 2x + 4 = -8 gives x = -6. The absolute value equation splits into two linear equations. Other options miscalculate one or both cases.
What is the solution to |x| = 0?
x = 0
x = -1
x = ±0
x = 1
Absolute value measures distance from x to zero, so |x| = 0 only when x = 0. ±0 is equivalent to 0 but is a misleading notation. Other options do not satisfy the equation.
Solve |x + 2| = 3.
x = 5 or x = -1
x = 1 or x = -5
x = -1 or x = -5
x = 1 or x = 5
Setting x + 2 = 3 yields x = 1, and x + 2 = -3 yields x = -5. The absolute value equation breaks into two cases. Other choices reflect sign errors.
What is the solution to |x - 1| = 0?
x = ±1
x = -1
x = 0
x = 1
Absolute value equals zero only when its argument is zero. Thus x - 1 = 0 gives x = 1. Other options do not satisfy the equation.
For the function y = |x|, what expression gives y when x < 0?
y = x
y = 0
y = -x
y = |x|
The definition of absolute value is y = x if x ? 0 and y = -x if x < 0, because distance is non-negative. For negative x, y = -x makes the output positive.
Evaluate |-7|.
|-7|
-7
7
0
Absolute value returns the non-negative magnitude, so |-7| = 7. Other options are incorrect values or restate the problem.
What is the solution set for the inequality |x| < 4?
0 ? x ? 4
x < -4 or x > 4
-4 < x < 4
-4 ? x ? 4
|x| < 4 means x lies less than 4 units away from zero, which gives all x strictly between -4 and 4. The OR and closed intervals in other options do not represent this strict inequality.
Solve the inequality |2x - 3| ? 5.
-4 ? x ? 1
x < -1 or x < 4
-1 ? x ? 4
x ? -1 or x ? 4
Split into -5 ? 2x - 3 ? 5, add 3 to each part to get -2 ? 2x ? 8, then divide by 2 yielding -1 ? x ? 4. Other choices either reverse the inequality or apply OR incorrectly.
What is the solution to |x + 1| > 2?
-3 < x < 1
x ? 1 or x ? -3
x < 1 or x > -3
x > 1 or x < -3
|x+1| > 2 splits to x+1 > 2 (x > 1) or x+1 < -2 (x < -3). The solution set is outside the interval [-3,1].
Where is the vertex of the function f(x) = |x - 2| + 3?
(2, 3)
(2, -3)
(-2, 3)
(3, 2)
The expression |x - 2| shifts the vertex from (0,0) to (2,0), and adding 3 moves it up to (2,3). This is the point where the graph changes direction.
Solve the equation |3x + 2| = |x - 4|.
x = 1 or x = -4
x = 4 or x = -2
x = -3 or x = 0.5
x = 3 or x = -0.5
Set 3x + 2 = x - 4 to get x = -3, and 3x + 2 = -(x - 4) to get x = 0.5. The other pairs do not satisfy the original equation.
How is the graph of y = |x + 1| shifted compared to y = |x|?
Down by 1 unit
Left by 1 unit
Right by 1 unit
Up by 1 unit
Adding inside the absolute value function causes a horizontal shift in the opposite direction, so y = |x + 1| moves the graph left by 1 unit.
Compute |2 - 5| + |5 - 2|.
9
0
6
3
|2 - 5| = 3 and |5 - 2| = 3, so their sum is 6. Absolute values make negative differences positive.
Solve |x - 1| + |x + 2| = 5.
x = -1 or x = 3
x = 3 or x = -2
x = 1 or x = -2
x = 2 or x = -3
Split into regions at x = -2 and x = 1, solving piecewise to get solutions x = 2 (x ? 1) and x = -3 (x < -2). The middle region yields no solution.
Find the solutions to |2x - 5| + |x + 3| = 10.
x = -4 or x = -2
x = 2 or x = -4
x = 4 or x = 2
x = 4 or x = -2
Consider regions determined by x = -3 and x = 2.5. Solving piecewise yields x = 4 and x = -2. Other values do not satisfy both absolute values.
What is the range of f(x) = |x - 3| + |x + 1| for real x?
[0, ?)
[2, ?)
[4, ?)
(0, ?)
The sum of distances to 3 and -1 is minimized at any x between these points, giving a minimum value of 4. Values increase without bound otherwise.
Solve the equation |x + 1| - |x - 2| = 3.
x > 2
No solution
x ? 2
x ? 2
For x ? 2, the expression becomes (x+1)-(x-2)=3, which holds for all x ? 2. The other regions do not satisfy the equation except at the boundary x = 2, which is included.
What is the minimum value of y = |x - 1| + |x - 4|?
2
3
5
4
Between x = 1 and x = 4 the distance sum equals (x-1)+(4-x) = 3. Outside this interval the sum grows larger.
Find the solution set of |x - 2| + |x + 1| ? 4.
(-?, ?)
[-1.5, ?)
[-1.5, 2.5]
(-?, 2.5]
Splitting into regions at x = -1 and x = 2 gives valid solutions from x = -1.5 up to x = 2.5. Other intervals do not satisfy the combined inequality.
Evaluate the limit as x ? ? of (|x + 1| - |x - 1|).
0
1
2
-2
For x > 1, both expressions drop their absolute signs: (x+1)-(x-1) = 2, so the limit is 2. Behavior for finite x does not affect the limit at infinity.
If f(x) = |x| + |x - 1| + |x - 2|, what is f(1)?
1
2
4
3
Compute each term: |1| = 1, |1 - 1| = 0, and |1 - 2| = 1, summing to 2.
True or False: The function g(x) = |x - a| is differentiable everywhere except at x = a.
True
False
Absolute value functions have a 'corner' at the point where the argument is zero, so g(x) = |x - a| is not differentiable at x = a. Everywhere else the derivative exists.
Solve the equation |x + 1| = |2x - 3|.
x = 4 or x = 0.666...
x = -4 or x = 1
x = 3 or x = 2
x = 4 or x = -0.666...
Set x + 1 = 2x - 3 to get x = 4, and x + 1 = -(2x - 3) to get x = 2/3. Both satisfy the original equation.
0
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Study Outcomes

  1. Interpret Absolute Value Concepts -

    Describe the definition and key properties of absolute functions, including distance interpretation and piecewise representation.

  2. Solve Absolute Value Equations -

    Apply algebraic techniques to find exact solutions of equations involving absolute value expressions.

  3. Solve Absolute Value Inequalities -

    Determine and graph the solution sets for both "less than" and "greater than" absolute value inequalities.

  4. Graph Absolute Value Functions -

    Sketch and analyze absolute value function graphs, identifying vertex, slope, and transformation parameters.

  5. Apply Problem-Solving Strategies -

    Use structured approaches to tackle practice absolute functions problems quiz and evaluate your understanding with instant feedback.

Cheat Sheet

  1. Definition and Piecewise Representation -

    Absolute value denotes distance from zero, so |x| = x if x ≥ 0 and |x| = - x if x < 0 (Stewart, Calculus). Remember the phrase "distance is always non-negative" to recall the rule. For example, |x - 2| = {x - 2 for x≥2; 2 - x for x<2} forms the basis of many math 144 quiz 1 problems.

  2. Solving |x - a| = b Equations -

    When |x - a| = b (b ≥ 0), split into x - a = b or x - a = - b, giving x = a ± b (Khan Academy). For instance, |x - 4| = 7 yields x = 11 or x = -3. Mastering this approach is key to crush your absolute value functions quiz.

  3. Tackling Absolute Value Inequalities -

    For |x - a| < b, rewrite as a - b < x < a+b; for |x - a| > b, write x < a - b or x > a+b (Purplemath). Example: |2x+1| ≤ 5 becomes - 5 ≤ 2x+1 ≤ 5, so - 1.5 ≤ x ≤ 2. Draw a number line to visualize solution intervals in your practice absolute functions problems quiz.

  4. Graphing and Transformations -

    Start with the parent graph y = |x| (a "V" with vertex at (0,0)) and apply y = |x - h| + k for shifts: right by h and up by k (MIT OpenCourseWare). So y = |x+3| - 2 moves the vertex to ( - 3, - 2). Visualizing these shifts helps you ace college algebra absolute functions tests.

  5. Isolation, Case-Split, and Check (I.C.E.) -

    Always Isolate the absolute value expression, perform Case-Split into positive/negative scenarios, then Evaluate and Check against the original equation (College Board). This I.C.E. mnemonic prevents extraneous roots. Confidence in this method will boost your performance on algebra absolute functions questions.

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