Question

What is the equation for the following function:
The shape of y = x^2 , but upside down, shifted right 3 units and up 4 units.

Answer 1

also, how did you solve it?

Answer 2

Step 1: make it upside down -> -x^2
Step 2: shifted right 3 units -> -(x - 3)^2
Step 3: shifted up 4 units -> 4 -(x - 3)^2

Answer 3

why do you also square the negative three?

Answer 4

I get the first and third steps, but not the second

Answer 5

Let me think of a good way to explain.

Answer 6

no prob. thanks :)

Answer 7

Think of it like this: to shift the graph of a function to the right by 3 units means to move each pair (x, y) in the graph to (x+3, y). Viewed another way, if (x, y) is a point in the new (shifted) graph, then it "came from" a point (x-3, y) in the original graph. Ok, this is not a very good explanation, but it may help you while I come up with a better one.

Answer 8

okay. i get why you move it over by three, but why do you square it?

Answer 9

why isn't it \[-x ^{2}-3\]

Answer 10

does that make any sense?

Answer 11

The "trick" goes like this: you know that the graph of the function y = x^2 has its lowest point at (x, y) = (0, 0). So if you shift it right by 3, it should have its lowest point at (3, 0). To obtain a value of y = 0, you need the rule of the function ("square it") to give you a zero when x = 3; so before you "square it", you need to subtract 3, so that then you tell "square it" to x-3 = 0, obtaining the desired value of y = 0.

Answer 12

It makes "sense", but it is incorrect.

Answer 13

k. that makes sense. Thanks so much!

Answer 14

Ok, here is an improved rewrite of my explanation. You are looking for a new function y' = f(x') such that to every point of its graph (x', y') corresponds a point (x, y) of the original graph such that

x' = x+3, and
y' = -y+4.

Notice that all three transformation (flip, shift right by 3, shift up by 4) are encoded in those two equations. Starting with the second equation and the fact that we know that y = x^2, we obtain

y' = -x^2 + 4

But we are not done yet, as our goal is to write y' as a function of x', not x; but then we use the first equation x' = x+3 to "substitute out x"

y' = -(x' - 3)^2 + 4

which is the desired answer. Notice the use of the parentheses to make sure we square x' - 3.

Hope this helps!