Question

What is the line of symmetry for the parabola whose equation is `y=2x^2-4x+1`?

Answer 1

`x=1`

Explanation:

Method 1: Calculus approach.

`y=2x^{2}-4x+1`
`\frac{dy}{dx}=4x-4`
The line of symmetry will be where the curve turns (due to the nature of the `x^{2}` graph.

This is also when the gradient of the curve is 0.

Therefore, let `\frac{dy}{dx}=0`

This forms an equation such that:
`4x-4=0`
solve for x, `x=1` and line of symmetry falls on the line `x=1`

Method 2: Algebraic approach.

Complete the square to find the turning points:

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

From this we can pick up the line of symmetry such that:
`x=1`