Question

How do you find the axis of symmetry and vertex point of the function: `y=2x^2-3x+2`?

Answer 1

Axis of symmetry is `x=3/4`, vertex is `(3/4, 7/8)`

Explanation:

Convert x terms into a perfect square as follows:

`y= 2(x^2 -(3x)/2 ) +2`

`y= 2(x^2 -(3x)/2 + 9/16 -9/16)+2`

`y=2(x^2-(3x)/2 +9/16) -9/8 +2`

`y=2(x-3/4)^2 +7/8`. This equation is now in the vertex form. Axis of symmetry is `x=3/4`, vertex is `(3/4, 7/8)`