SAT 2 math help question ???

Question

If f(x)= ax^2 + bx + c, how must a and b be related so that the graph of f(x-3) will be symmetric about the y-axis?

A. a=b

B. b=0, a is any real number

C. b=3a

D. b=6a

E. a=1/9 b

it would be great if you could explain....how many methods can you solve this ....show all of them ...thanks

Answer 1

If  you want answer, skip to last paragraph (or line) If you want method (how i found out, read the whole d**n thing, and where I screwed up, sorry)

I am a lazy a** so I probably won't give everything, but hopefully enough.  (or not- I just finished writing this)  Anyways, symmetric about y-axis obviously means just that, which means the vertex of the equation must be at the y-axis.  However, the question calls for the graph which is shifted to the right by 3, such that you have to compensate.  Since the original vertex would be at x=0 (at y-axis), then original B would be equal to zero.  However, cuz it is shifted, it ain't.  (i.e. f (x) = x^2 - 9 would be an example of original, modified is f(x-3) = (x-3)^2 - 9)

Taking all of that into consideration, best thing to do is to solve out f(x-3), which gives

a(x-3)^2 + bx - 3b + c

= ax^2 - 6ax + 9a + bx - 3b +c

since you are looking at the y-axis, C does not matter, so you can eliminate that number (make it equal to zero), giving you:

ax^2 - 6ax + 9a + bx - 3b

Now, going back, you would want this equation to have a vertex at x=0, such that with c equal to zero, at x = 0, the equation is equal to zero.  However, in this equation that would not work since you would be left with 0=0.  So, new idea.  I gonna do a bit of algebra here, can't explain, sorry

ax^2 - 6ax + 9a + bx - 3b = 0

a(x^2 - 6x + 9) + b(x-3) = 0

(x - 3)[a(x - 3) + b] = 0

a(x-3) + b = 0

ax - 3 + b = 0

ax + b = 0

let x = 0

a(0) + b = 0

b = 0, thus a = any number

which would give you answer (B)

this does not work out, because for the example x^2 - 9 = f(x), f(x-3) would give x^2 - 6x, which goes to x(x-6) = 0, so that vertex is at 6.  

but u notice, x^2 would have vertex at 0, and so u need eliminate -6x, so set 6x = b(x-3).  

gives 6x = bx - 3b

=bx = 9x

b = 9

when A=1 (in this example)

thus, answer is (E),  = 1/9B