Question

The function f(x)=x^2. the graph of g(x) is f(x) translated to the right 3 units and down 3 units. What is the function rule for g(x)?

Answer 1

to move a function up or down is easy — you just add the appropriate value to the result of the function. to move down 3 units, we add -3 to f(x).

Answer 2

to move right or left, we have to adjust the number we put into the function.

Answer 3

Up and down you add to the equation
Left and right you you add to the exponent.

Answer 4

@nelsonjedi not the exponent, the function argument

Answer 5

So is the function rule the equation? or is it the rules of moving the equation?

Answer 6

so if we add 3 to the argument of the function, the result will be the same as if we had computed the function 3 units to the right. for example, f(0+3) = f(3), so whatever value the function normally has at x = 3 will now appear at x = 0. does that make sense?

Answer 7

Wouldn't it be the other way around? whatever appears at x=0 would appear at x=3

Answer 8

Sorry my bad. Yes the function arguement

Answer 9

adding to the function argument shifts the graph to the left. subtracting from the function argument shifts the graph to the right.

Answer 10

No, make yourself a graph and try it. use f(x) = x^2. here's a graph of x^2 vs (x+3)^2

Answer 11

the blue graph is x^2, the purple is (x+3)^2. at x = -3, the purple graph = 0, which is what the blue graph = at x = 0.

Answer 12

Okay. so would my answer be g(x)=3x^2-3

Answer 13

so if we wanted that same parabola to have its vertex over at (5,0) instead of the (0,0) that we get with \(f(x) = x^2\) we have to subtract 5 from the argument to \(f(x)\), right?

Answer 14

or g(x)=3+x^2-3

Answer 15

No. You've got the up/down part correct, but multiplying x^2 by 3 just changes the scale up and down...

This graph shows the original in blue, and the shift down three units (gotten by adding -3 to the result of f(x))

Answer 16

Like I said, that part is easy. If we want g(x) to just be a shift f(x) down 3 units, we just say \[g(x) = f(x)-3\]

Answer 17

To shift f(x) right or left, we have to do something like
\[g(x) = f(x-a)\]

Answer 18

and to do both right/left and up/down, we'll have something like
\[g(x) = f(x-a) + b\]

Answer 19

because \(f(x) = x^2\] that will become
\[g(x) = (x-a)^2+b\]

Answer 20

so the equation would be g(x)=f(x-3)+3 or
g(x)=f(x-3)-3

Answer 21

the latter one, yes.

Answer 22

Okay. thanks

Answer 23

sorry about messing up the formatting a post ago, that should have been

because \(f(x) = x^2\) that will become
\[g(x) = (x-a)^2+b\]

(a and b can be negative or positive, of course)

Answer 24

And that would be my answer to the original question?

Answer 25

well, probably you should expand it out.
\[g(x) = f(x-3)-3 = \]can you expand that, given that \(f(x) = x^2\)

Answer 26

expand it how?

Answer 27

well, replace f(x-3) with its expanded form: how would you compute the value of f(x-3)?

Answer 28

what you have now is in fact a correct answer, but you should understand how to complete the process because that may be what is desired here

Answer 29

what you've written is math-ese for "make g(x) by translating the graph of f(x) to the right 3 units and down 3 units" but it doesn't tell us what g(x) actually is so that we could graph it, do you see what I mean?

Answer 30

distribute the f into the parenthesis?

Answer 31

if \[f(x) = x^2\]\[f(x+1) = (x+1)^2\]right?

Answer 32

just like \[f(3) = (3)^2\]

Answer 33

Yes

Answer 34

okay, so if \(g(x) = f(x-3)-3\), what does \(g(x)=\) in terms of \(x\), no \(f()\) allowed?

Answer 35

It would be the some problem without the f, right?

Answer 36

\[g(x) = f(x-3)-3\]replace \( f(x-3)\) with \((x-3)^2\)
\[g(x) = f(x-3)-3 = (x-3)^2-3\]and now expand that out

Answer 37

Expand it?

Answer 38

multiply it out

Answer 39

\[g(x) = (x-3)^2-3 =(x-3)(x-3)-3 = x^2-3x-3x+9-3\]\[ = x^2-6x+6\]

Answer 40

now we have an equation for \(g(x)\) that we can use without needing to know anything about \(f(x)\)

Answer 41

but if you plug in a bunch of values and plot it, you'll see that in fact, it is the graph of f(x) shifted 3 units to the right and 3 units down

Answer 42

Wow, okay. I didn't know you could do that just by multiplying it out. Thank you.

Answer 43

Yes, \[(x-3)^2 = (x-3)(x-3)\]

Answer 44

Right.

Answer 45

think of the \(f(x)\) notation as a recipe for how to construct the function.

if you have something like \(f(x) = x^3+2x^2+5x-3\) that means you can construct the formula for \(f(a)\) by cubing a, adding 2a^2, adding 5a, and subtracting 3, for whatever value a is. if a happens to be some ugly polynomial with 17 terms, well, it's going to be a lot of work, but the approach is the same :-)

Answer 46

Okay. I wish I could look at math like that. I can never seem to see how that pieces fit together. Thanks. :D

Answer 47

and the \[g(x) = f(x-3) -3\]notation works for any function \(f(x)\)! Say we had a different parabola, \(f(x) = x^2+2x-4\)

If we wanted to shift it 3 left and 3 down,
\[g(x) = f(x-3) -3 = (x-3)^2+2(x-3)-4 -3\]\[=x^2-6x+9+2x-6-4-3 = x^2-4x-4\]\[g(x) = x^2-4x-4\]

And here they both are, graphed together:

Answer 48

anyhow, the key takeaway from this exercise is:

to shift a function up or down by \(a\), add or subtract from the function result:
\[f(x) \pm a\]

to shift a function left or right by \(a\), add or subtract from the function argument:
\[f(x\pm a)\]

Answer 49

Okay. I got it, I think. haha. thanks

Answer 50

you'll have many opportunities to test your understanding of this in the coming years :-)

Answer 51

I hope not too much in college.