The graph of the f(x) is show below. Graph each transformed function and list in words the transformation used?

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I'm kind of confused on how to shift the graphs. I do know that you use this formula
y=A+(B(x−h))+k
to graph these functions.

1 Answer
Nov 29, 2017

This is the original graph of #f(x)#.

enter image source here

a) This is the graph of #g(x)=f(-x)+1#.
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b) This is the graph of #h(x)=-1/2f(x)-2#.
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c) This is the graph of #k(x)=2f(x+3)-1#.
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Explanation:

Use this equation to solve algebraically: #y=af(b(x-h))+k#.

For the graph of #g(x)=f(-x)+1#, there is a reflection in the #y#-axis and a vertical translation 1 unit up. In other words, #b# is negative, and #k = 1#.

#(x,y)# on #y=f(x)# will be #(-x,y+1)# on #g(x)=f(-x)+1#.

#(-1,4) ->(1,5)#
#(1,0) ->(-1,1)#
#(3,4)->(-3,5)#

For the graph of #h(x)=-1/2f(x)-2#, there is a reflection in the #x#-axis, a vertical compression by a factor of 1/2, and a vertical translation of 2 units down. In other words, #a=-1/2#, and #k=-2#.

#(x,y)# on #y=f(x)# will be #(x,1/2y-2)# on #h(x)=-1/2f(x)-2#.
#(-1,4) ->(-1,-4)#
#(1,0) ->(1,-2)#
#(3,4)->(3,-4)#

For the graph of #k(x)=2f(x+3)-1#, there is a vertical stretch by a factor of 2, a horizontal translation 3 units left, and a vertical translation 1 unit down. In other words, #a=2#, #h=-3#, and #k=-1#.

#(x,y)# on #y=f(x)# will be #(x-3,2y-1)# on #k(x)=2f(x+3)-1#.

#(-1,4) ->(-4,7)#
#(1,0) ->(-2,-1)#
#(3,4)->(0,7)#