Math problems? about parabolas?

Question

Uhh.. how do you do this?
The graph of 5x2squared shifted 8 units to the right and 8 units below the axis??
OR!!!
The graph of y = 7x2squared 10 units above the origin?

Answer 1

OK. For the first part, I'm not sure if you mean (5x)^2 or 5(x)^2. So..
If it's y=(5x)^2, let y'= your new function
To shift the graph 8 units to the right and 8 down:
y'=(5(x-8))^2 - 8
If it's y=5(x)^2, .....
y'=5(x-8)^2 - 8
(** To sort of explain this, when you plug in 8 for x in the new function y', you get (5(8-8))^2 -8. This equals (5(0))^2 -8= -8. So,while the vertex used to be (0,0) for the function y, now it is (8, -8) for y'. This change is the same for all points in y ~what used to be (1,25) or (1,5) are now (9, 17) and (9, -3) **)

For the second part, to shift the graph 10 units above the origin, you just have to add 10 to it. So, since I don't now if you mean (7x)^2 or 7(x)^2, if y=(7x)^2, then:
y'=(7x)^2 + 10
if y= 7(x)^2, then:
y'=7(x)^2 + 10

I hope this helps. It's hard trying to explain math without a pencil and paper. Good luck!