Question

How do you find the vertex and the intercepts for `y= x^2 - 4x + 5`?

Answer 1

Vertex `(2,1)`
Y-intercept `(0, 5)`

Explanation:

Given -

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

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

At `x=2`

`y=2^2-4(2)+5=4-8+5=9-8=1`

Vertex `(2,1)`

At `x=0`

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