## The problem

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What is the vertex ordered pair?

Axis of symmetry is x=

The domain is

The range is

## Answer provided by our tutors

**Vertex of the quadratic equation**

For a given quadratic y = ax^2 + bx + c, the vertex (h, k) is found by computing

h = –b/2a,

*k* = (4* ac* –

*2)/4*

*b^*

*a*y=2(x - 3)^2 - 2

If we expand:

y = 2x^2 - 12x + 16

a = 2

b = -12

c = 16

h = -(-12)/(2*2)

h = 3

k = (4*2*16 - (-12)^2)/(4*2)

k = -2

The vertex is (3, -2).

**Axis of symmetry** is x = 3

**The domain** is the entire set of real numbers R.

**The range** is [-2, +~).