Question

Find the vertex and axis of symmetry of parabola `y=2x^2+4x-1`?

Answer 1

`x=-1`

Explanation:

graph{2x^2+4x-1 [-32.48, 32.47, -16.24, 16.22]}

The equation for the axis of symmetry of a quadratic is `x="x-value of the vertex"`

To find the vertex (with a standard form quadratic equation) use the formula:

`(-(b))/(2(a))="x-value of the vertex"`

In this case, plugin your `a` and your `b` to find the x-value:

`(-(4))/(2(2))`

`(-4)/4=-1`

So the `"x-value"` is `-1`, so your equation for the axis of symmetry is:

`x=-1`

Answer 2

Axis of symmetry is `x+1=0` or `x=-1`

Explanation:

Vertex form of equation of a parabola (read quadratic equation of form `y=ax^2+bx+c`), is `y=a(x-h)^2+k`,
where `(h,k)` is the vertex and
axis of symmetry is `x-h=0 or x=h,`

As `y=2x^2+4x-1`

or `y=2(x^2+2x)-1`

= `2(x^2+2x+1-1)-1`

= `2(x+1)^2-2-1`

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

Hence, axis of symmetry is `x+1=0` or `x=-1`

And vertex is `(-1,-3)`