Question

How do you find x-intercepts, axis of symmetry, maximum or minimum point, and y-intercept for the equation `y = x^2+2x-3`?

Answer 1

To find `x` intercepts, find `x` to make `y=)`.
That is: solve `x^2+2x-3 = 0`
`(x-1)(x+3) = 0`
The `x` intercepts are `-3` and `1`.

Find `y` intercepts by finding `y` when `x=0`.
`y = (0)^2 +2(0) -3 = -3`
The `y` intercept is `-3`.

Find the vertex:

The vertex of `y = ax^2 +bx +c` has `x` coordinate `-b/(2a)`
So the graph of `y=x^2+2x-3` has vertex at `x= -(2)/(2(1)) = -1`
The axis of symmetry is the line: `x=-1`

Because `a=1` is positive, the parabola opens upward, so there is a minimum value for `y`. That minimum occurs where `x = -1`, so the minimum is: `y = (-1)^2 +2(-1) -3 = 1-2-3= -4`
The minimum value of `y` is `-1`.

I don't know what you've been taught to call the "minimum point", but I'll assume it has 2 coordinates, so the "minimum point" is the vertex.
I expect the "minimum point" to be `(-1, -4)`.