Question 178080
{{{y = 2(x-3)(x+1)}}}
The x-intercepts are the roots. The roots are values of {{{x}}} that make
{{{Y=0}}}. Looking at the equation, if {{{x=3}}}
{{{y = 2(3-3)(x+1)}}}
{{{0 = 2*0*(x+1)}}}
And if {{{x=-1}}}
{{{y = 2(x-3)(-1+1)}}}
{{{0 = 2(x-3)*0}}}
So, {{{x=3}}} and {{{x=-1}}} must be the x-intercepts.
The vertex is always exactly midway between the roots, so
{{{x[v] = (-1 + 3)/2}}}
{{{x[v] = 2/2}}}
{{{x[v] = 1}}}
So, the vertex is at ({{{1}}},{{{y[v]}}})
{{{y = 2(x-3)(x+1)}}}
{{{y[v] = 2(1-3)(1+1)}}}
{{{y[v] = 2*(-2)*2}}}
{{{y[v] = -8}}}
So, the vertex must be at (1,-8)
Now I'll graph the equation to prove what I said
{{{y = 2(x-3)(x+1)}}}
{{{y = 2*(x^2 - 2x - 3)}}}
{{{y = 2x^2 - 4x - 6}}}
{{{ graph( 500, 500, -10, 10, -10, 10, 2x^2 - 4x - 6) }}}