Question

How do you graph `y=x^2- 6x+2`?

Answer 1

`"y-int" = 2`
`"x-int" = 3+-sqrt7`
vertex = `(3,-7)`

Explanation:

`y=x^2- 6x+2`

your function is in the form: `ax^2+bx +c`

`a=1`
`b=-6`
`c=2`

From this we can derive:

y-intercept = c

y-int = 2

x-intercept(s) if they exist are the solutions or roots:

`y=x^2- 6x+2`

Roots (found with the quadratic formula): `3+-sqrt7`

axis of symmetry is: `aos=(-b)/(2a) = (-(-6))/(2*1)=3`

vertex is: `(aos, "*"f(aos))`
*this means you put the aos value back into the function as x and solve for y.

vertex is: `(3, 3^2- 6*3+2)`

vertex is: `(3,-7)`

Finally here is your graph:

graph{x^2- 6x+2 [-9.84, 10.16, -7.48, 2.52]}