Question

How do you find the x intercepts for the parabola and a vertex `y = x^2 – 6x + 5`?

Answer 1

Your quadratic is in the form `y=ax^2+bx+c` where:
`a=1`
`b=-6`
`c=5`

To find the `x` intercept(s) set `y=0` and solve for `x` the 2nd degree equation: `x^2-6x+5=0`

The coordinates of the vertex are:
`x_v=-b/(2a)`
`y_v=-Delta/(4a)`
Where `Delta=b^2-4ac`