Question

How do you graph the parabola `y= 2x^2 - 4x + 5` using vertex, intercepts and additional points?

Answer 1

Y intercept `(0, 5)`
Vertex `(1,3)`

Explanation:

Given-

`y=2x^2-4x+5`

At `x=0`

`y=2(0)^2-4(0)+5`
`y=5`

Y-Intercept `(0, 5)`

Vertex-

`x=(-b)/(2a)=(-(-4))/(2xx2)=4/4=1`

At `x=1`

`y=2(1)^2-4(1)+5=2-4+5=3`

Take a range of values for `x` in such a way they include `0` to `1`

Table

Graph