A standard function form for translations and dilations is y=a(x-h)2+c where:
a is the vertical dilation (sometimes we use y=k(x-h)2+c)
h is the x-coordinate of the vertex
c is the y-coordinate of the vertex
Vertex (h,c)
To find the equation when y=x2 is shifted 3 units down and 4 units right, remember y=a(x-h)2+c where a is the vertical dilation, h is the x-coordinate of the vertex, and c is the y-coordinate of the vertex.
The vertex in y=x2 (which can also be written as y=(x-0)2+0) is (0,0). If you shift the vertex 3 units down (downward movement along the y-axis) and 4 units right (shift to the left along the x-axis) you will get a new vertex of (0+4,0+(-3))=(4,-3)
To get the new equation, substitute the new vertex coordinates (4,-3) into y=(x-0)2+0.
This gives y=(x-4)2-3 which simplifies to y=(x-4)2-3.
y=(x-4)2-3
Question 2 of 5
2. Question
Find the equation when (x-1)2+(y-3)2=1 is reflected about the y axis and then shifted down 2 units.
Reflections about the y-axis means that we have to replace xโ-x.
A standard circle equation is in the form: (x-h)2+(y-c)2=r2
h is the x-coordinate for the centre
c is the y-coordinate for the centre
First we find the equation when (x-1)2+(y-3)2=1 is reflected about the y-axis, we replace the x by -x. We get (-x-1)2+(y-3)2=1 or simply (x+1)2+(y-3)2=1.
Remember (x-h)2+(y-c)2=r2 where h is the x-coordinate of the centre, and c is the y-coordinate of the centre.
The center for (x+1)2+(y-3)2=1 is (-1,3).
Next we will shift this circle 2 units down along the y-axis. Since the formula indicates (y-c)2 it becomes (y--c)2โ(y+c)2.
Notice it is +c even though we are going downwards. So as a rule of thumb for this formula: (x-h)2+(y-c)2=r2 anytime -c or -h is inside the brackets we move in the opposite direction.
This gives (x+1)2+(y-1)2=1.
The centre is (-1,1).
(x+1)2+(y-1)2=1
Question 3 of 5
3. Question
Find the equation when y=ex is vertically dilated by a factor of 4 and shifted 5 units to the left.
A standard exponential function form for translations and dilations is
k is the vertical dilationy=ke(x+h)
h is the x-coordinate for the horizontal translation
To find the equation when y=ex is vertically dilated by a factor of 4 and shifted 5 units left, remember y=ke(x+h) where k is the vertical dilation and h is the horizontal translation.
The equation y=ex when vertically dilated by a factor of 4 becomes y=4ex. Then applying the translation of 5 units to the left
(left means +5), y=4ex becomes y=4ex+5.
y=4ex+5
Question 4 of 5
4. Question
Find the equation when y=2x2-3x is shifted 1 unit left and then reflected about the y axis.
Reflections about the y-axis means that we have to replace xโ-x.
A standard function for a horizontal translation is y=(x+h) where +h is a left shift along the x-axis.
To find the equation when y=2x2-3x is shifted 1 unit left. Now, in order to shift it 1 unit to the left we replace x with (x+1).
y
=
2(x+1)2-3(x+1)
y
=
2(x2+2x+1)-3x-3
y
=
2x2+4x+2-3x-3
y
=
2x2+x-1
Then we reflect the equation y=2x2+x-1 about the y-axis by replacing x with -x. The equation becomes, y=2(-x)2+(-x)-1 or simply y=2x2-x-1.
y=2x2-x-1
Question 5 of 5
5. Question
Find the transformed version of y=x3 when you have a horizontal dilation factor of 12, a horizontal translation of 3 units left, and a vertical dilation of 4 are applied.
The application of the horizontal dilation (factor) on a x variable is xfactor.
A standard function form for a horizontal translation is y=f(x+h) where +h is a shift to the left movement along the x-axis.
A standard function form for a vertical dilation is kf(x) where k is the vertical dilation.
To transform y=x3 with a horizontal dilation factor of 12, horizontal translation of 3 units left, and a vertical dilation of 4, start by applying the horizontal dilation factor of 12 first. Do this by using xfactor and factor=12.
y=
(x12)3
Apply the horizontal dilation factor of 12. Remember xfactor.
=
(2x)3
Simplify
=
(2(x+3))3
Apply the horizontal translation of 3 units left. Use y=f(x+3).
=
(2(x+3))3
Simplify
Now apply the vertical dilation of k=4. Use kf(x).