A standard function form for translations and dilations is y=a(x-h)2+cy=a(xโh)2+c where:
aa is the vertical dilation (sometimes we use y=k(x-h)2+cy=k(xโh)2+c)
hh is the xx-coordinate of the vertex
cc is the yy-coordinate of the vertex
Vertex (h,c)(h,c)
To find the equation when y=x2y=x2 is shifted 33 units down and 44 units right, remember y=a(x-h)2+cy=a(xโh)2+c where aa is the vertical dilation, hh is the xx-coordinate of the vertex, and cc is the yy-coordinate of the vertex.
The vertex in y=x2y=x2 (which can also be written as y=(x-0)2+0)y=(xโ0)2+0) is (0,0)(0,0). If you shift the vertex 33 units down (downward movement along the yy-axis) and 44 units right (shift to the left along the xx-axis) you will get a new vertex of (0+4,0+(-3))=(4,-3)(0+4,0+(โ3))=(4,โ3)
To get the new equation, substitute the new vertex coordinates (4,-3)(4,โ3) into y=(x-0)2+0y=(xโ0)2+0.
This gives y=(x-4)2-3y=(xโ4)2โ3 which simplifies to y=(x-4)2-3y=(xโ4)2โ3.
y=(x-4)2-3y=(xโ4)2โ3
Question 2 of 5
2. Question
Find the equation when (x-1)2+(y-3)2=1(xโ1)2+(yโ3)2=1 is reflected about the yy axis and then shifted down 22 units.
Reflections about the yy-axis means that we have to replace xโ-xxโโx.
A standard circle equation is in the form: (x-h)2+(y-c)2=r2(xโh)2+(yโc)2=r2
hh is the xx-coordinate for the centre
cc is the yy-coordinate for the centre
First we find the equation when (x-1)2+(y-3)2=1(xโ1)2+(yโ3)2=1 is reflected about the yy-axis, we replace the xx by -xโx. We get (-x-1)2+(y-3)2=1(โxโ1)2+(yโ3)2=1 or simply (x+1)2+(y-3)2=1(x+1)2+(yโ3)2=1.
Remember (x-h)2+(y-c)2=r2(xโh)2+(yโc)2=r2 where hh is the xx-coordinate of the centre, and cc is the yy-coordinate of the centre.
The center for (x+1)2+(y-3)2=1(x+1)2+(yโ3)2=1 is (-1,3)(โ1,3).
Next we will shift this circle 22 units down along the y-yโaxis. Since the formula indicates (y-c)2(yโc)2 it becomes (y--c)2โ(y+c)2.(yโโc)2โ(y+c)2.
Notice it is +c+c even though we are going downwards. So as a rule of thumb for this formula: (x-h)2+(y-c)2=r2(xโh)2+(yโc)2=r2 anytime -cโc or -hโh is inside the brackets we move in the opposite direction.
This gives (x+1)2+(y-1)2=1.(x+1)2+(yโ1)2=1.
The centre is (-1,1)(โ1,1).
(x+1)2+(y-1)2=1(x+1)2+(yโ1)2=1
Question 3 of 5
3. Question
Find the equation when y=exy=ex is vertically dilated by a factor of 44 and shifted 55 units to the left.
A standard exponential function form for translations and dilations is
kk is the vertical dilationy=ke(x+h)y=ke(x+h)
hh is the xx-coordinate for the horizontal translation
To find the equation when y=exy=ex is vertically dilated by a factor of 44 and shifted 55 units left, remember y=ke(x+h)y=ke(x+h) where kk is the vertical dilation and hh is the horizontal translation.
The equation y=exy=ex when vertically dilated by a factor of 44 becomes y=4exy=4ex. Then applying the translation of 55 units to the left
(left means +5+5), y=4exy=4ex becomes y=4ex+5y=4ex+5.
y=4ex+5y=4ex+5
Question 4 of 5
4. Question
Find the equation when y=2x2-3xy=2x2โ3x is shifted 11 unit left and then reflected about the yy axis.
Reflections about the yy-axis means that we have to replace xโ-xxโโx.
A standard function for a horizontal translation is y=(x+h)y=(x+h) where +h+h is a left shift along the x-axis.
To find the equation when y=2x2-3xy=2x2โ3x is shifted 11 unit left. Now, in order to shift it 11 unit to the left we replace xx with (x+1)(x+1).
yy
==
2(x+1)2-3(x+1)2(x+1)2โ3(x+1)
yy
==
2(x2+2x+1)-3x-32(x2+2x+1)โ3xโ3
yy
==
2x2+4x+2-3x-32x2+4x+2โ3xโ3
yy
==
2x2+x-12x2+xโ1
Then we reflect the equation y=2x2+x-1y=2x2+xโ1 about the yy-axis by replacing xx with -xโx. The equation becomes, y=2(-x)2+(-x)-1y=2(โx)2+(โx)โ1 or simply y=2x2-x-1y=2x2โxโ1.
y=2x2-x-1y=2x2โxโ1
Question 5 of 5
5. Question
Find the transformed version of y=x3y=x3 when you have a horizontal dilation factor of 1212, a horizontal translation of 33 units left, and a vertical dilation of 44 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).