Find the vertex of the quadratic equation y= 1/2(x-2)^2-1

Question

Consider the quadratic equation y=1/2(x-2)²-1

1: what is the vertex

2:find the axis of symmetry

3:describe in words the transformations

4: find a point to the right of the vertex

5:find a 2nd point to the right of the vertex

6:state the domian and range

7: graph the quadratic

Answer 1

You have a quadratic in the form of:

y = (x-h)^2 + k

 

1) The vertex is (h,k) or (2,-1)

2) The axis of symmetry is x=h  (or x=2)

3) The a=1/2 widens the parabola and is + so the graph opens up

The h = +2 shifts the parabola 2 to the right of the y axis

The k = -1 shifts the parabola 1 down from the x axis

4) Use a value > h to substitute in for x and find the y of a point to the right of the vertex

5) Use another value > h to substitute in for x and find the y of a point to the right of the vertex

7) to help graph it

Use a value < h to substitute in for x and find the y of a point to the left of the vertex