Question

How to graph a parabola `y=x^(2)-2x-15`?

Answer 1

You have two choices
1. complete the square to get standard form, find vertex and 2 other points.
2. Use the equation `x=(-b/2a)` to find the x-coordinate of the vertex.

Explanation:

Complete the square...
`y=(x^2-2x+(b/2)^2)-15-(b/2)^2`
`y=(x^2-2x+1)-15-1`
`y=(x-1)^2-16` so the vertex is at (1, -16)
find two more points, use x-values near the vertex.
(0, -15) and by symmetry (2, -15) and (-1, -12).

Alternatively,
Use `h=(-b/2a)` for the x-coordinate of the vertex
and `k=` the y-value when you substitute x = h into the equation.
write as `y=a(x-h)^2+k` here a=1, the vertex is (h, k).
You need to calculate 2 more points as above.