Question

How do you find its vertex, axis of symmetry, y-intercept and x-intercept for `f(x) = x^2-2x-3`?

Answer 1

Find vertex of f(x) = x^2 - 2x - 3

Explanation:

x of vertex: `x = (-b/2a) = 4/2 = 2`

y of vertex:`y = f(2) = 4 - 4 - 3 = -3`

x of axis of symmetry = x of vertex = 2

To find y-intercept, make x = 0 -> y = -3

To find x-intercepts, make `f(x) = x^2 - 2x - 3 = 0`

Since (a - b + c = 0), then one real root is (-1) and the other
is `(-c/a = 3).`
x-intercepts: -1 and 3