Problem 1 :
What is the vertex: y = 3|x - 1| + 2
a) (1, -2) b) (1, 2)
Solution :
y = 3|x - 1| + 2
By comparing the given absolute value function with
y = a|x - h| + k
Vertex (h, k):
(1, 2)
Problem 2 :
What is the vertex: y = 2|x|
a) (2, 0) b) (0, 0)
Solution :
y = 2|x|
By comparing the given absolute value function with
y = a|x - h| + k
Vertex (h, k):
(0, 0)
Problem 3 :
What is the vertex: y = |x| + 5
a) (5, 0) b) (0, 5)
Solution :
y = |x| + 5
By comparing the given absolute value function with
y = a|x - h| + k
Vertex (h, k) :
(0, 5)
Problem 4 :
The graph of y = -2|x + 1| - 3 is
a) Vertically stretched b) Vertical shrink
Solution :
Comparing the given equation with y = a |x - h| + k
a = -2, since the sign of a is negative, it is reflection and the value of a is 2.
a > 1, it is vertically shrink.
Problem 5 :
The graph of y = -3/5|x + 3| + 10 is
a) Vertically stretched b) Vertical shrink
Solution :
Comparing the given equation with y = a |x - h| + k
a = -3/5, since the sign of a is negative, it is reflection and the value of a is 3/5.
0 < a < 1, it is vertically stretch.
Problem 6 :
The graph of y = 15|x| is
a) Vertically stretched b) Vertical shrink
Solution :
Here a = 15
a > 1, it is vertically shrink.
Problem 7 :
The graph of y = 5/3|x + 2| - 1 is
a) Vertically stretched b) Vertical shrink
Solution :
a = 5/3
a > 1, it is vertically shrink.
Problem 8 :
The graph of y = 5/3|x + 2| - 1
a) Opens up b) Opens down
Solution :
y = a|x - h| + k
y = 5/3|x + 2| - 1
a = 5/3
It is positive. So opens up.
Problem 9 :
The graph of y = -2|x + 1| - 3
a) Opens up b) Opens down
Solution:
y = a|x - h| + k
y = -2|x + 1| - 3
a = -2
It is negative. So opens down.
Problem 10 :
The graph of: y = 3|x| - 2 is
a) Translated vertically b) Translated horizontally
Solution :
Comparing the given function y = 3|x| - 2 with the parent function y = |x|, for k we have -2. So, the graph is moved towards down 2 units. So translated vertically is the answer.
Problem 11 :
The graph of: y = 3|x| - 2 is
a) Translated left 2 b) Translated down 2
Solution :
Since the value of k is -2, we have to do translation 2 units down.
Problem 12 :
In order for the graph to be a vertical shrink, what will be the value of a
a) -1 < a < 1 b) a > 1
Solution :
If a > 1, then it is vertical shrink.
Problem 13:
Domain of: y = -2|x + 1| - 3
a) (-∞, ∞) b) -∞ < x < 1
Solution :
There is no restriction for x values for the function given above, so
Domain: (-∞, ∞) x € R
All real numbers are domain.
Problem 14 :
Range of: y = 3|x| - 1
a) (-∞, 3] b) [-1, ∞),
Solution :
y = 3|x| - 1
Range: [-1, ∞), y ≥ -1
Problem 15 :
Range of: y = -|x| - 2
a) [-2,-∞) b) [-2, ∞)
Solution :
y = -|x| - 2
Range: [-2,-∞), y ≤ -2
Problem 16 :
Range of: y = -|x - 5| + 4
a) (-∞, 4] b) [4, ∞)
Solution :
y = -|x - 5| + 4
Range: [4, ∞), y ≥ 4
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