CHARACTERISTICS OF ABSOLUTE VALUE FUNCTIONS WORKSHEET

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Find the following and graph it.

(i) Vertex

(ii)  Slope

(iii) y-intercept and x-intercept

(iv) domain and range

(v)  Increasing and decreasing interval.

Problem 1 :

f(x) = -3β”‚x - 4β”‚ + 3

Solution

Problem 2 :

f(x) = -1/2β”‚x - 2β”‚ + 4

Solution

Problem 3 :

f(x) = β”‚x - 3β”‚ - 2

Solution

Problem 4 :

f(x) = 3β”‚xβ”‚

Solution

Problem 5 :

y = 3/5 β”‚xβ”‚- 6

Solution

Problem 6 :

y =β”‚x + 4β”‚

Solution

Answer Key

1) 

2) 

3) 

4) 

5)

6) 

Graph the following absolute value function by finding the following.

(i) Vertex

(ii)  Slope

(iii)  Direction of opening

(iv) x and y intercepts

(v) Domain and range

(vi) Increasing and decreasing

Problem 1 :

y = 3|x - 3|

Solution

Problem 2 :

y = -|x| + 4

Solution

Problem 3 :

y = (4/3) |x + 2| - 5

Solution

Problem 4 :

y = -(3/2) |x - 3| + 2

Solution

Problem 5 :

Match each function with its graph. Explain your reasoning.

i. f(x) = β”‚x + 2β”‚ βˆ’ 2

ii. g(x) = βˆ’β”‚x βˆ’ 2β”‚ + 2

iii. f(x) = βˆ’β”‚x βˆ’ 2β”‚ βˆ’ 2

iv. m(x) = β”‚x + 2β”‚ + 2

graphing-abs-function-q1

Solution

Answer Key

1)  Vertex is at (3, 0)

Slope (a) = 3

Direction of opening = open up

x-intercept is at (3, 0)

y-intercept is at (0, 9).

  • All real values is domain.
  • Range is 3 β‰€ y β‰€ βˆž
  • To the left of minimum, it is decreasing.
  • To the right of minimum, it is increasing.

2)  Vertex is at (0, 4).

x-intercepts are (4, 0) and (-4, 0).

y-intercept is at (0, 4).

Slope (a) = -1

The curve will open down.

  • All real values is domain.
  • Range is 4 ≀ y β‰€ -∞
  • To the left of maximum, it is increasing.
  • To the right of maximum, it is decreasing.

3)  Vertex is at (-2, -5).

x-intercept is at (7/4, 0) and (-19/2, 0).

y-intercept (0, -7/3).

Slope (a) = 4/3

The curve will open up.

  • All real values is domain.
  • Range is 4 ≀ y β‰€ -∞
  • To the left of minimum, it is decreasing.
  • To the right of minimum, it is increasing.

4)  Vertex is at (3, 2).

x-intercept is at (13/3, 0) and (5/3, 0).

y-intercept is (0, -5/2).

Slope (a) = -3/2

The curve will open down.

  • All real values is domain.
  • Range is -5 ≀ y β‰€ ∞
  • To the left of maximum, it is increasing.
  • To the right of maximum, it is decreasing.

5) i) Graph C is correct

ii) Graph B is correct.

iii) Graph D is correct.

iv) Graph A is correct.

Solve for x using

i) graphical method    ii) an algebraical method.

Problem 1 :

|x + 2| = 2x + 1

Solution

Problem 2 :

Consider the graph, which shows y1 = |6 - 2x| and y2 = 18. Use the graph and use algebraic properties to solve |6 - 2x| = 18.

solving-absolute-value-function-graphing-q1

Solution

Problem 3 :

You are asked for the year of the Emancipation Proclamation in the United States on a test. The correct answer is 1863. You guessed g and you were off by 4 years. What equation’s solution gives the possible values of g?

a) |1863 - 4| = g      b) |g| = 1863 - 4     c) |g - 1863|  = 4    d) |g - 4|  = 1863

Solution

Problem 4 :

Determine whether the number is a solution to the equation

60 = |n - 90| 

a) 30     b) βˆ’30    c) 150     d) βˆ’150

Solution

Problem 5 :

 Use the table below to solve each sentence.

a) |2x - 3| = 7     b) |2x - 3| < 7     c) |2x - 3| > 7

solving-absolute-value-function-graphing-q2.png

Solution

Answer Key

1)

absofxplus2p1

the solution is (1, 3).

Solving algebraically :

|x + 2| = 2x + 1

x + 2 = (2x + 1)

x + 2 = 2x + 1

x - 2x = 1 - 2

-x = -1

x = 1

(x + 2) = -(2x + 1)

x + 2 = -2x - 1

x + 2x = -1 - 2

-3x = 3

x = -1

2) the point of intersections are -6 and 12.

3) |g - 1863|  = 4

4) 30 and 150 are solutions.

5) (-∞, -2) (5, ∞)

Write the equation of the graph. Then give its range as an inequality.

Problem 1 :

equationformgraphofabq1

Solution

Problem 2 :

equationformgraphofabq2

Solution

Problem 3 :

equationformgraphofabq3

Solution

Problem 4 :

equationformgraphofabq4

Solution

Problem 5 :

equationformgraphofabq5

Solution

Problem 6 :

equationformgraphofabq6

Solution

Problem 7 :

Describe and correct the error in graphing the function.

graphing-asbsolute-value-function-q1

Solution

Problem 8 :

Describe and correct the error in graphing the function.

graphing-asbsolute-value-function-q2.png

Solution

Answer Key

1) 

y = 1 |x +1| + 2

Range is 2 β‰€ y β‰€ βˆž

2)

y = -1 |x + 3| - 2

Range is -2 β‰€ y β‰€ -∞

3) 

y = -2 |x - 2| + 3

Range is 3 ≀ y β‰€ -∞

4)

y = (1/2) |x + 2| - 3

Range is -3 ≀ y β‰€ ∞

5)

y = (-3/4) |x - 1|

Range is 1 ≀ y β‰€ -∞

6)

 

y = (2/3) |x + 2|

Range is -2 ≀ y β‰€ ∞

7) By observing the graph, the function opens up but the vertex is at (-1, -3) and that is the error.

8) By observing the graph, it opens up. But it should be open down and that is the error.

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