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

Jun 9, 2015

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

#### Explanation:

x of vertex: $x = \left(- \frac{b}{2} a\right) = \frac{4}{2} = 2$

y of vertex:$y = f \left(2\right) = 4 - 4 - 3 = - 3$

x of axis of symmetry = x of vertex = 2

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

To find x-intercepts, make $f \left(x\right) = {x}^{2} - 2 x - 3 = 0$

Since (a - b + c = 0), then one real root is (-1) and the other
is $\left(- \frac{c}{a} = 3\right) .$
x-intercepts: -1 and 3