Question

What is the axis of symmetry and vertex for the graph `y = 2x^2 - 8x + 4`?

Answer 1

Complete the square (or use `(-b)/(2a)`)

Explanation:

To complete the square for `y=2x^2-8x+4`:
First take out the 2 for the first two terms
`y=2(x^2-4x)+4`
Then take the value for b (which is 4 here), divide by 2 and write it like this:
`y=2(x^2-4x+2^2-2^2)+4`
They both cancel out each other so adding these two terms to the equation isn't a problem.
Within your new equation take the first term and third term (`x^2` and 2) inside the brackets and put the sign of the second term (`-`) between these two so it looks something like this:
`y=2((x-2)^2-2^2)+4`
Then simplify:
`y=2(x-2)^2-4`

The x coordinate of the vertex is found by taking the expression within the brackets and simply doing:
`0=x-2`
so
`x=2`
and the y coordinate is the number behind the brackets.
`y=-4`

So the coordinates of the vertex becomes:
`(2, -4)`
And the axis of symmetry:
`x=2`

Another way to get the same answer is to use `(-b)/(2a)`
`x=(-b)/(2a)`
`x=8/(2(2))`
`x=2`

and substitute 2 in into `y=2x^2-8x+4` to find `y`.