Question

How do you use the important points to sketch the graph of `y=x^2-4x+7`?

Answer 1

The graph intersects the y-axis at `(0, 7)` and the vertex is at `(2, 3)`
and the term `y` carries a + sign, therefore the parabola opens upward.

Explanation:

Solving for the vertex by "Completing the square method"

`y=x^2-4x+7`
`y=x^2-4x+4-4+7`
`y=(x-2)^2+3`
`y-3=(x-2)^2`
`(x-2)^2=y-3` the vertex form showing the vertex at `(2, 3)`

Solving for the intercepts
`y=x^2-4x+7`
when `x=0`
`y=0^2-4*0+7`
`y=7`

`(0, 7)` is a point on the parabola and the y-axis

The graph of `y=x^2-4x+7`
graph{y=x^2-4x+7[-20,20,-10,10]}

God bless....I hope the explanation is useful.