How do you find the vertex and the intercepts for #y= -2(x+3)(x-1)#?

So intercepts are when the parabola touches the axes.

It is when one variable 0, thus, we can sub in #0# and solve for the other variable.

X-intercepts

That is the zeros AKA roots AKA solutions. The equation is in factored form, giving us the zeros without having to calculate it. Thus, the x-intercepts are #(-3, 0)# and #(1, 0)#.

Y-intercept

We can find the #y"-intercept"# in two ways: changing the equation into standard form, OR sub #x# as #0#. We'll do the easier method, subbing in.

#y=-2(x+3)(x-1)#

#y=-2(0+3)(0-1)#

#y=-2(3)(-1)#

#y=6#

Therefore, the #y"-intercept"# is #(0, 6)#.

Vertex

To find vertex in factored form, the easiest method is to find the axis of symmetry, and sub that in as #x# and solve for #y#.

The axis of symmetry can be calculated given the formula: #x=(r+s)/2#.
=> #r# and #s# are the zeros.
=> #x# is the axis of symmetry AKA the #x"-component"# in the vertex.

Finding AoS

#x=(r+s)/2#

#x=(-3+)/2#

#x=-2/2#

#x=-1#

Subbing in AoS to find y-component of vertex

#y=-2(x+3)(x-1)#

#y=-2(-1+3)(-1-1)#

#y=-2(2)(-2)#

#y=8#

Therefore, the vertex is #(-1, 8)#.

Let's check our work by graphing it.

graph{-2(x+3)(x-1) [-10, 10, -0.24, 9.76]}

As you can see, the vertex and intercepts are correct.

Hope this helps :)