You have not understood the question.. We do NOT know that f(x) is symmetric about the Y-axis. It could be, but that is of no consequence.
The question is this f(x-3) = a(x-3)^2 + b(x-3) + c -- Now make THAT symmetric about the Y-axis.
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?
The book says to take -b/2a (x coordinate of f(x)). Since the new graph is 3 to the right, -b/2a = -3. So b = 6a.
My question is how do we know that the original f(x) is symmetrical about the y-axis???
You have not understood the question.. We do NOT know that f(x) is symmetric about the Y-axis. It could be, but that is of no consequence.
The question is this f(x-3) = a(x-3)^2 + b(x-3) + c -- Now make THAT symmetric about the Y-axis.
I don't mean to sound redundant. but how do we know that taking the x coordinate of the vertex and setting it equal to -3 will make THAT symmetric about the y-axis? I get that the graph is shifted three to the right, but three to the right of what?
What is the name of your book? If the x co-ordinate of the vertex is -3, then there is no way to get symmetry around the y axis. The x co-ordinate of the vertex must be zero, only then can you have symmetry around the y axis. This means THE COEFFICIENT OF x must work out to be zero in a(x-3)^2 + b(x-3) + c. It does not matter what b is in the ax^2 + bx + c.
You have to expand it out, combine alike terms and identify the coefficient of x. This is the new b. Zero must be the new coefficient of x. It must equal zero in order to have symmetry around the y axis.
ax^2 - 6ax + 9a + bx - 3b + c
-6a + b = 0
so b = 6a (this is the old b)
just like book said.
I guess if you do this many times then you see the shortcut for this form, which is what book gave, unfortunately, shortcuts don't explain WHY, for that you need to break things down and challenge your understanding.