Question

What is the axis of symmetry and vertex for the graph `y=2x^2+6x+4`?

Answer 1

Vertex is `(-1/2,-3/2)` and axis of symmetry is `x+3/2=0`

Explanation:

Let us convert the function to vertex form i.e. `y=a(x-h)^2+k`,

which gives vertex as `(h,k)` and axis of symmetry as `x=h`

As `y=2x^2+6x+4`, we first take out `2` and make complete square for `x`.

`y=2x^2+6x+4`

= `2(x^2+3x)+4`

= `2(x^2+2xx3/2xx x+(3/2)^2)-(3/2)^2xx2+4`

= `2(x+3/2)^2-9/2+4`

= `2(x-(-3/2))^2-1/2`

Hence, vertex is `(-1/2,-3/2)` and axis of symmetry is `x+3/2=0`
graph{2x^2+6x+4 [-7.08, 2.92, -1.58, 3.42]}