The problem asks us to find the axis of symmetry and the vertex of the graph of the given quadratic functions. Also, we are asked to sketch the graph, but since I cannot draw, I will not do it. (1) $y = 2(x-1)^2$ (2) $y = -\frac{1}{2}(x+2)^2$

1. Problem Description

The problem asks us to find the axis of symmetry and the vertex of the graph of the given quadratic functions. Also, we are asked to sketch the graph, but since I cannot draw, I will not do it.

(1) $y = 2(x-1)^2$

(2) $y = -\frac{1}{2}(x+2)^2$

2. Solution Steps

(1) The quadratic function is given by $y = 2(x-1)^2$.

The vertex form of a quadratic function is given by $y = a(x-h)^2 + k$, where $(h,k)$ is the vertex and the axis of symmetry is $x=h$.

In this case, we have $y = 2(x-1)^2 + 0$, so $a=2$, $h=1$, and $k=0$.

Therefore, the vertex is $(1,0)$ and the axis of symmetry is $x=1$.

(2) The quadratic function is given by $y = -\frac{1}{2}(x+2)^2$.

We can rewrite this as $y = -\frac{1}{2}(x-(-2))^2 + 0$.

In this case, we have $a=-\frac{1}{2}$, $h=-2$, and $k=0$.

Therefore, the vertex is $(-2,0)$ and the axis of symmetry is $x=-2$.

3. Final Answer

(1) Vertex: $(1,0)$, Axis of symmetry: $x=1$

(2) Vertex: $(-2,0)$, Axis of symmetry: $x=-2$