Question 837764
There is a role for algebraic manipulations based on that "foiling method", or a "completing the square" method", depending on what form of the function they give you. However, I am guessing that your problem does not require that. (I have to guess because I do not know what functions or graphs you were given).
Because I am guessing what you need to know, I will cover many possibilities and the answer will be long.
 
When teachers talk about "transform" and "parent function" to graph a function,
you have to use either
A) your brain, or
B) "recipes" from the book or from the teacher
to change the known graph of the "parent function" into a similar graph for the function given, or to match a given graph.
 
From the "parent function" {{{y=x^2}}} you can derive many related functions. Their graphs look similar but moved ("translated" or "shifted") left or right, and/or up or down, and/or "stretched" or "shrunken" horizontally or vertically, and/or "flipped".
I'll give you examples, and what my brain tells me about them. They may remind you of "recipes" and "rules" you got from school and books.
When you have to "describe" or "explain", it is better to follow the teacher's (or book's) favorite wordings and recipes, because independent use of your brain may not be understood, or accepted.
 
The graph of {{{y=x^2}}} looks like this:
{{{drawing(300,300,-5,5,-1,9,
grid(1),graph(300,300,-5,5,-1,9,x^2)
)}}}
 
SHIFTING/TRANSLATING VERTICALLY:
If you wanted the graph of {{{y=x^2+3}}} you would just move each point in that graph 3 units up, so both functions graphed on the same axes would look like this:
{{{drawing(300,300,-5,5,-1,9,
grid(1),graph(300,300,-5,5,-1,9,x^2,x^2+3),
blue(circle(-1,1,0.1)), blue(circle(-1,4,0.1)),
blue(arrow(-1,1.1,-1,3.9)),
blue(circle(0.5,0.25,0.1)), blue(circle(0.5,3.25,0.1)),
blue(arrow(0.5,0.35,0.5,3.15)),
blue(circle(2,4,0.1)), blue(circle(2,7,0.1)),
blue(arrow(2,4.1,2,6.9))
)}}} (Just for illustration, I marked 3 of the points, and drew the arrows showing how they change from one graph to the other. Your teacher most likely does not want to see the arrows, or the marked points). All the points in the graph move the same way. The whole graph moves (is "translated" or "shifted") 3 units up.
Similarly, the graph of {{{y=x^2-2}}} is like the graph of {{{y=x^2}}} , but with all points moved ("translated" or "shifted") 2 units down.
 
SHIFTING/TRANSLATING HORIZONTALLY:
If you add a number to the {{{x}}} before applying the function, you have a different effect.
The graph for {{{y=x^2}}} has {{{y=0}}} for {{{x=0}}}, {{{y=1}}} for {{{x=1}}},
and {{{y=4}}} for {{{x=2}}} .
The graph for {{{y=(x-3)^2}}} would have {{{y=0}}} for {{{x=3}}}, where {{{x-3=0}}} .
It would have {{{y=1}}} for {{{x=4}}}, where {{{x-3=1}}} .
It would have {{{y=4}}} for {{{x=5}}}, where {{{x-3=2}}} .
All the points would be moved 3 units to the right because you have {{{(x-3)}}} where you had just {{{x}}} :
{{{drawing(300,300,-3,7,-1,9,
grid(1),graph(300,300,-3,7,-1,9,x^2,(x-3)^2),
blue(circle(1,1,0.1)), blue(circle(4,1,0.1)),
blue(arrow(1.1,1,3.9,1)),
blue(circle(0,0,0.1)), blue(circle(3,0,0.1)),
blue(arrow(0.1,0,2.9,0)),
blue(circle(2,4,0.1)), blue(circle(5,4,0.1)),
blue(arrow(2.1,4,4.9,4))
)}}} (Just for illustration, I marked 3 of the points, and drew the arrows showing how they change from one graph to the other. Your teacher most likely does not want to see the arrows, or the marked points). All the points in the graph move the same way. The whole graph moves (is "translated") 3 to the right.
Similarly, the graph of {{{y=(x+1)^2}}} is like the graph of {{{y=x^2}}} , but with all points moved ("translated" or "shifted") 1 unit to the left.
 
STRETCHING AND SHRINKING VERTICALLY:
A factor in front of the square has a different effect.
The graph of {{{y=2x^2}}} is like the graph of {{{y=x^2}}} , but "vertically stretched", because for each {{{x}}} , {{{y=2x^2}}} assigns a {{{y}}} value twice as great as {{{y=x^2}}} :
{{{drawing(300,300,-5,5,-1,9,
grid(1),graph(300,300,-5,5,-1,9,x^2,2x^2),
blue(circle(1,1,0.1)), blue(circle(1,2,0.1)),
blue(arrow(1,1.1,1,1.9)),
blue(circle(2,4,0.1)), blue(circle(2,8,0.1)),
blue(arrow(2,4.1,2,8.1))
)}}}
Something like that would happen for any other factor greater than 1.
For factors between 0 and 1, the graph would look vertically "shrunken", as shown below for {{{t=x^2}}} and {{{y=0.2*x^2)}}} .
{{{drawing(300,300,-5,5,-1,9,
grid(1),graph(300,300,-5,5,-1,9,x^2,0.2x^2)
)}}}
 
FLIPPING (also called "reflecting"):
A minus sign in front of the function flips {{{red(y=x^2)}}} over the x-axis turning it into {{{green(y=-x^2)}}} as shown below:
{{{graph(300,300,-5,5,-25,25,x^2,-x^2)}}}
The "transformation" could be called "flipping" or "reflecting" over the x-axis.
Putting a minus sign in front of the function can be seen as multiplying the whole function times {{{(-1)}}} , since {{{-x^2=(-1)*x^2}}} .
Any negative factor, would flip and at the same time stretch or shrink the graph vertically, as shown below for {{{red(y=x^2)}}} , {{{green(y=-2x^2)}}} , and {{{blue(y=-0.2x^2)}}}
{{{graph(300,300,-5,5,-25,25,x^2,-2x^2,-0.2x^2)}}}
 
OTHER CHANGES:
You could talk about stretching and shrinking horizontally, and about flipping over the y-axis. That would involve applying a factor to the {{{x}}} before applying the function.
However, with {{{y=x^2}}} (squaring) as the "parent function", I would avoid those moves.
 
COMBINATIONS:
You could do several transformations to the function in a certain order.
Sometimes the same result can be obtained in more that one way.
 
EXAMPLES:
Say you are given the graph below. How was {{{y=x^2}}} transformed to get that?
{{{drawing(300,300,-7,3,-10,10,
grid(1),graph(300,300,-7,3,-10,10,-3(2x+8)^2+8) )}}} It was flipped upside down, so there is a minus sign up front.
The graph looks stretched vertically, so there is a factor up front with an absolute value greater than 1.
The vertex has been moved from (0,0) 4 units to the left, and 8 units up to (-4,8).
A shift 4 units to the left means that what was squared is {{{(x+4)}}}.
A shift up by 8 units means there is a {{{"+ 8"}}} at the end.
What about the vertical stretch?
For {{{y=x^2}}} , points where {{{x}}} is 1 unit away from the vertex at (0,0) have
{{{y=1^1=1}}} , 1 unit away from the {{{y=0}}} at the vertex.
This function, for {{{x}}} 1 unit away from the vertex  we are at {{{x=-4-1=-5}}} or at {{{x=-4+1=-3}}} . At those points we seem to have {{{y=-4}}} , so we are {{{8-(-4)=12}}} units away from the {{{y=8}}} at the vertex, so {{{12}}} (with the minus sign for the flip) is my factor, and I can write the function as
{{{y=-12(x+4)^2+8}}}
I would say that from {{{y=x^2}}} the function has been
flipped (or reflected) over the x-axis,
vertically stretched by a factor of 12,
shifted (or moved, or translated) 4 units left, and
shifted (or moved, or translated) 8 units up.
 
Say you are given the function {{{green(y=-2(x+1)^2+3)}}} .
The transformations from {{{red(y=x^2)}}} are listed below, followed by the graphs.
The {{{x+1}}} that is squared means a 1 unit shift to the left.
The {{{-2}}} at front means a vertical stretch by a factor of 2, and an upside down flip.
The {{{"+ 3"}}} at the end means a shift up by 3 units.
{{{drawing(300,300,-7,3,-3,7,
grid(1),graph(300,300,-7,3,-3,7,x^2,-2(x+1)^2+3) )}}}
 
If my function were {{{y=5(2x+3)^2+8}}} , I would use algebra to re-write it a way I like better:
{{{y=5(2x+3)^2+8}}} --> {{{y=5(2(x+3/2))^2+8}}} --> {{{y=5*2^2(x+3/2)^2+8}}} --> {{{y=20(x+3/2)^2+8}}} .
Then I would say the transformations from {{{y=x^2}}} are:
a {{{3/2=1.51}}} unit shift to the left,
a vertical stretch by a factor of 20, and
a shift up by 8 units.
Someone else may say that there was a horizontal shrinking by a factor of 2 and a vertical stretching by a factor of 5, but those two changes compound into the vertical stretch by a factor of 20 I listed. It is different ways to state the same thing.
 
IF MORE ALGEBRA KNOWLEDGE IS TO BE USED:
If someone gave me something like {{{y=4(x-2)(x-3)}}} I might think of foiling,
but I would not need to do that.
I know that, with zeros (x-intercepts) at {{{x=2}}} and {{{x=3}}} ,
the vertex and axis of a symmetry of that quadratic function is halfway between the zeros, at {{{x=2.5=5/2}}}.
At that point, {{{y=4(2.5-2)(2.5-3)=4*0.5*(-0.5)=4*(-0.25)=-1}}} .
That, and the {{{4}}} in front, tells me that {{{y=3(x-2)(x-3)}}} is {{{y=x^2}}} transformed by
a {{{2.5=5/2}}} unit shift right,
a {{{1}}} unit shift down, and
a vertical stretch by a factor of {{{4}}} .
 
If someone was mean enough to give me something like {{{y=3x^2-15x+10}}} , I would "complete the square" as follows.
{{{y=3x^2-15x+10}}}
{{{y=(3x^2-15x)+10}}}
{{{y=3(x^2-5x)+10}}}
{{{y=3(x^2-5x+25/4-25/4)+10}}}
{{{y=3(x^2-5x+25/4)-3*25/4+10}}}
{{{y=3(x^2-5x+25/4)-75/4+10}}}
{{{y=3(x^2-5x+25/4)-35/4}}}
{{{y=3(x-5/2)^2-35/4}}}
That function is {{{y=x^2}}} transformed by
a {{{2.5=5/2}}} unit shift right,
a {{{35/4=8.75}}} unit shift down, and
a vertical stretch by a factor of {{{3}}} .