Question

Help, please. （〃・ω・〃）

A video game producer is using the quadratic function f(x) = 4 – x2 to model the path of an arrow shot by a game player. The path of the arrow is shown in the graph below. The point (2, 3) represents the top of the castle wall.

http://prntscr.com/37xtxb

If the player moves so that the arrow will clear the wall at the maximum height, what is the new function that the game producer would need to use? Use complete sentences and explain your work.

Answer 1

Okay, can you see, from an objective and lay-man perspective, that we just need to shift that parabola (trajectory path) one unit to the right?

Answer 2

Yes, I can. (>v<)

Answer 3

So, all this requires is a horizontal shift of one unit to the right.
I suppose explaining that in complete sentences is something that comes naturally to you? :)

Answer 4

Hehehehe. Ohmigosh, that was it? (OvO)

Answer 5

Not quite.
You still need the new equation, silly :P

Answer 6

Do you know how to apply a horizontal shift onto functions?

Answer 7

~gasp~

I think I might know how to do this...

Answer 8

Try it. Tell me what you get.

Answer 9

http://media.tumblr.com/tumblr_m5odx0Mg9b1qcwic6.gif

\[f(x)+1=4-x ^{2}\]or\[f(x+1)=4-x ^{2}\]

Answer 10

~confident~

Answer 11

<sigh>

Answer 12

Serious work needs to be done with you XD

Answer 13

Okay.... first off,
f(x) + 1
is NOT the same as f(x+1)

got it? XD

Answer 14

I know that, I was just wondering which of the two it is....

Answer 15

Go figure. Which do you think? (If any at all, that is :P)

Answer 16

I think it is the f(x + 1). :3

Answer 17

gah...
yes... technically, it is correct
but might I inquire as to how you got that answer?

Answer 18

Yeah. >v>

Answer 19

Well, how DID you arrive at f(x+1) = bla bla bla

It's highly unorthodox.
As in... nobody I know does it like that....

Answer 20

Um...because that is how they taught me when I learned about functions...when I learned about vertical and horizontal shifts...

Answer 21

I see.
Well yeah, as I said, it's technically correct (okay... it's correct) but I prefer it like this...
to shift n-units to the right, simply replace all the x's (except the one in f(x) ) with (x - n)

Like, if you want to shift 1 unit to the right, then replace x with x-1:
\[\Large f(x) = 4 - x^2\]\[\Large f(x) = 4 - \color{blue}{(x-1)}^2\]

Note that the new function (the one below) is equivalent to your f(x+1) thingy... to demonstrate:
\[\Large f(\color{red}x) = 4 - (\color{red}x - 1)^2\]
\[\Large \begin{array} f (\color{red}{x+1}) &=& 4 - (\color{red}{x+1}-1)^2\\&=&4 - x^2\end{array}\]

So...yeah
You got it... just in a way I find really really REALLY weird.

:P

Well done anyway, I take it you can do the rest... ^_^

Answer 22

\[\Large \begin{array} \ f(\color{red}{x+1}) &=& 4 - (\color{red}{x+1}-1)^2\\&=&4 - x^2\end{array}\]
sorry this is what I meant...