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)