Thread: Finding the vertex and x-intercepts of an equation

How do i find the vertex and the x-intercepts of y=2(x-3)(x+1)?

Originally Posted by scorpio

How do i find the vertex and the x-intercepts of y=2(x-3)(x+1)?

The equatio describes a parabola opening up.

You find the x-intercepts if y = 0. Since you have the equation in factored form you can read the values of the x-intercepts directly: x = 3 or x = -1

Expand the brackets and rewrite the equation in vertex-form:

y = 2(x² - 2x - 3) = 2(x²-2x+1 - 1 - 3) = 2(x-1)² - 8

Therefore the vertex has the coordinates: V(1, -8)