Can you see if my answer is right?

Identify the transformations of the graph of f(x)=-3(x+3)^2-3 that would cause the graph's image to have a vertex (3,3). then write the transformed function.

I got f(x)=(x-3)^2+3. Shift right 6 and up 6. Is this right?

Question

Can you see if my answer is right?

Identify the transformations of the graph of f(x)=-3(x+3)^2-3 that would cause the graph's image to have a vertex (3,3). then write the transformed function.

I got f(x)=(x-3)^2+3. Shift right 6 and up 6. Is this right?

anonymous · Answer

*answers with nothing but pointless letters*

anonymous · Answer

Sort of right.  f(x) = 3*(x-3)^2+3 has the vertex at (3,3) and also has the same parabolic shape as the original equation.

anonymous · Answer

AND your proposed solution removed the negative sign from the (x-3)^2 term, so the parabola is inversed.

anonymous · Answer

a better solution is -3(x-3)^2+3
because it is created by subtracting -6 from x and -6 from f(x), i.e. f(x) - 6 = -3(x-6+3)^2-3, the simplify into f(x) = -3(x-3)^2+3
you get the same curvature and concaveness of the original equation, just shifted up 6 and over 6.