Question 18332
Once your quadratic equation is in the standard form:{{{ax^2 + bx + c]= 0}}}, the x-coordinate of the vertex is given by {{{-b/2a}}}.

{{{-b/2a = -(-4)/2(1)}}} = {{{4/2 = 2}}}

Now find the y-coordinate of the vertex by substituting this value of x (x=2) into the quadratic equation and solve for y.

{{{y = (2)^2 - 4(2) - 2}}}
{{{y = 4 - 8 - 12}}}
{{{y = -16}}}

The vertex is at (2, -16)

The x-intercepts can be found by solving the quadratic equation.

{{{x^2 - 4x - 12 = 0}}} Solve by factoring.
{{{(x+2)(x-6) = 0}}} Apply the zero products principle.
{{{x+2 = 0}}} and/or {{{x-6 = 0}}}
If x+2 = 0, then x = -2
If x-6 = 0, then x = 6
The x-intercepts are: (-2, 0) and (6, 0)

See the graph below:
{{{graph(300,200,-10,10,-20,10,x^2-4x-12)}}}