FIND X AND Y INTERCEPTS OF ABSOLUTE VALUE FUNCTION WORKSHEET
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Find x and y intercepts of following absolute functions :
Problem 1 :
y = 3|x - 1| + 2
Problem 2 :
y = 2|x|
Problem 3 :
y = |x| + 5
Problem 4 :
y = -2|x + 1| - 3
Problem 5 :
y = -3/5|x + 3| + 10
Problem 6 :
y = 15|x|
Problem 7 :
y = 5/3|x + 2| - 1
Problem 8 :
y = -2|x + 1| - 3
Problem 9 :
You ride your bicycle around a circular trail one time. The function
f(x) = (−1/3) ∣x − 4.5∣ + 1.5
represents the shortest distance (in miles) along the trail between you and your starting point after x minutes.
(a) Graph the function. Find the domain and range in this context.
(b) Interpret the intercepts and the vertex.
Answer Key
1) no x-intercept, y -Intercept is (0, 5).
2) x -Intercept is (0, 0), y -Intercept is (0, 0).
3) no x-intercept, y -Intercept is (0, 5).
4) no x-intercept, y -Intercept is (0, -5)
5) x-intercepts are (41/3, 0) and (-59/3, 0), y -Intercept is (0, 41/5)
6) x -Intercept is (0, 0), y -Intercept is (0, 0).
7) x -Intercept are (-7/5, 0) and (-13/5, 0), y -Intercept (0, 7/3).
8) no x-intercept, y -Intercept is (0, -5).
9) the domain is {x | 0 ≤ x ≤ 9} and the range is { f (x) | 0 ≤ f (x) ≤ 1.5}
vertex (4.5, 1.5)
Identify the vertex.
- Determine if the graph opens up or down.
- Determine if the graph has a maximum or minimum and its value.
- Decide if the graph is narrower, wider, or the same width as the parent graph.
Problem 1 :
y = -|x + 1|
Problem 2 :
y = 7|x - 3| - 4
Problem 3 :
y = -2/3|x - 1|
Problem 4 :
y = 5/2|x + 9| - 1
Problem 5 :
y = 3/4|x + 3| - 6
Problem 6 :
y = -|x| + 5
Problem 7 :
You are running a ten-mile race. The function d(t) = (1/8) ∣t − 40∣ represents the distance (in miles) you are from a water stop after t minutes.
a. Graph the function. Find the domain and range in this context.
b. Interpret the intercepts and the vertex. When is the function decreasing? increasing? Explain what each represents in this context.
Problem 8 :
Write the vertex of the absolute value function f(x) = ∣ax − h∣ + k in terms of a, h, and k.
Problem 9 :
Describe the transformation from the graph of f to the graph of g.
Answer Key
1) Vertex (-1, 0)
a = -1, the curve will open down.
The maximum is at x = -1.
width of this curve will be same.
2) Vertex (3, -4)
a = 7, the curve will open up.
The minimum is at x = 3.
a = 7 which is greater than 1. So, width of this curve will be narrower.
3) Vertex (1, 0)
a = -2/3, the curve will open down.
The maximum is at x = 1
a = 2/3 < 1, so it is wider.
4) Vertex (-9, -1)
a = 5/2, the curve will open up.
The minimum is at x = -9
a = 5/2 > 1, so the curve is narrower.
5) Vertex (-3, -6)
a = 3/4, the curve will open up.
The minimum is at x = -3
a = 3/4 < 1, so it is wider.
6) Vertex (0, 5)
a = -1, the curve will open down.
Maximum is at x = 0
a = 1. so it is same.
7)
domain is t ≥ 0
the range is 0 ≤ d(t) ≤ 5
x-intercept is at (40, 0) and y-intercept is (0, 5)
Since the absolute value function opens up, decreasing 0 < t < 40 and increasing at 40 < t ≤ 80
Distance covered in between 40 seconds is lesser than distance covered in between 40 to 80 seconds.
8) Vertex is at (h/a, k)
9) the value of k is -3.
To graph absolute value function, we have to find the following characteristics of the given function.
(i) Find vertex
(ii) x - intercepts (roots, zeroes, solutions) and y - intercept
(iii) Slope and Reflections (or) Direction of opening
(iv) Domain and Range
(v) Increasing/decreasing interval
Graph the following absolute value function :
Problem 1 :
y = 3|x - 3|
Problem 2 :
y = -|x| + 4
Problem 3 :
y = (4/3) |x + 2| - 5
Problem 4 :
y = -(3/2) |x - 3| + 2
Problem 5 :
Match each function with its graph. Explain your reasoning.
i. f(x) = │x + 2│ − 2
ii. g(x) = −│x − 2│ + 2
iii. f(x) = −│x − 2│ − 2
iv. m(x) = │x + 2│ + 2
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 ≤ y ≤ ∞
- 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 ≤ -∞
- 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 ≤ -∞
- 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 ≤ ∞
- 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.
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