Graph Transformation and finding the Vertex and Axis of Symmetry

Question

I have only one attempt left for this quiz. I would greatly appreciate it if someone could verify if my answers are correct or help me if my answers are wrong.

For question 1a), these are my following answers:

1. vertex: (0,2)
2. axis of symmetry: x=2
3. y-intercept: (0,2)

For question 1b), I have no idea how to solve it.

1. Equation of new curve, y = -1/2(4x^2+2)+3
2. vertex(?,?)

Would greatly appreciate if anyone could help me and provide a brief explanation. Thanks!

Answer 1

The axis of symmetry should be $x = 0$ instead.

Let $y = f(x)$ be the original.

Applying transformation #1 to $f(x)$, we get $h(x) = 0.5f(x)$.

Applying transformation #2 to $h(x)$, we get $k(x) = -0.5h(x)$

Applying transformation #3 to $k(x)$, we get the resultant $j(x) = k(x) + 3$

Eliminating $f(x), h(x), k(x)$ to get $j(x) = –0.5 (4x^2 + 2) + 3$ as described.

Either (1) re-write $j(x)$ in the form $ax^2 + bx + c$ to get the vertex located at $(– b/[2a], j(– b/[2a]))$.

Or (2) use “completing the square” method (or some other more simpler ways) to re-write $j(x)$ in the form $A(x – p)^2 + q$. Then, the vertex is at $(p, q)$.