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Question 1 of 5
Given x2+(y+2)2=9
Reflect x2+(y+2)2=9 across the
x-axis
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Reflections around the x-axis have the property y=-f(x).
Replace y→-y.
To sketch x2+(-y+2)2=9, we first look at the graph of x2+(y+2)2=9.
Since the standard form of the equation of a circle is (x-h)2+(y-k)2=r2, we know that the center of the circle is (h,k) and the radius is r.
From the equation of the original function x2+(y+2)2=9, the center of the circle is (0,-2) and the radius is r=√9=3.
Sketch the graph of the original function x2+(y+2)2=9 using the center and the radius.
Reflect across the x-axis by replacing y for -y, therefore x2+(-y+2)2=9. Which is the same as
x2+(y-2)2=9.
Sketch the graph of x2+(-y+2)2=9 by following the shape and radius of the original graph.
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Question 2 of 5
Given y=f(x)
Sketch y=f(-x)
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Reflections around the y-axis have the property y=f(-x).
Replace x→-x.
To sketch y=f(-x), we first look at the graph of y=f(x).
Use a table of values to find at least four points on the function y=f(x).
Sketch the graph of y=f(x) using the table of values.
Reflect the points marked for y=f(x) across the y-axis.
Sketch the graph of y=x2+1 by following the shape of the original graph but connecting the new translated points.
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Question 3 of 5
Given y=f(x)
Sketch y=-f(-x)
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Reflections around the y-axis have the property y=f(-x).
Replace x→-x.
Reflections around the x-axis have the property y=-f(x)
Replace y→-y.
To sketch y=-f(-x), we first look at the graph of y=f(x).
Use a table of values to find at least four points on the function y=f(x).
x |
-4 |
0 |
4 |
8 |
9 |
y |
4 |
0 |
4 |
0 |
-3 |
Sketch the graph of y=f(x) using the table of values.
Reflect the points marked for y=f(x) across the x-axis.
y=-f(-x) consists of one reflection about the x-axis and then one reflection about the y-axis.
This is exactly the same as a 180° rotation about the origin.
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Question 4 of 5
Given y=f(x)
Sketch y=-f(x)
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Reflections around the x-axis have the property y=-f(x).
Replace y→-y.
To sketch y=-f(x), we first look at the graph of y=f(x).
Use a table of values to find at least four points on the function y=f(x).
Sketch the graph of y=f(x) using the table of values.
Reflect the points marked for y=f(x) across the x-axis.
Sketch the graph of y=-f(x) by following the shape of the original graph but connecting the new translated points.
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Question 5 of 5
Given f(x)=1x-3
Sketch the graph of y=f(-x)
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Reflections around the y-axis have the property y=f(-x).
Replace x→-x.
To sketch the transformed function y=f(-x), start by sketching the original function f(x)=1x-3.
Find the asymptotes of the function f(x)=1x-3.
x-3= |
|
0 |
Set the denominator x-3 equal to 0. |
x= |
|
3 |
Add 3 to both sides of the equation. |
Then, sketch the original function f(x)=1x-3 by using the Asymptotex=3. Remember that this function is a hyperbola.
Now sketch the transformed function y=f(-x) which is f(x)=1-x-3=-1x+3.
Find the asymptotes of the function f(x)=-1x+3.
x+3= |
|
0 |
Set the denominator x+3 equal to 0. |
x= |
|
-3 |
Subtract 3 to both sides of the equation. |
Sketch the transformed function f(x)=-1x+3 by using the Asymptotex=-3. Remember that this function is a hyperbola.