Question 991544
{{{y=4(x+1)(x-1)}}}
Since the zeros are at {{{x=-1}}} and {{{x=1}}}, then the vertex will be at {{{x=0}}}.
When {{{x=0}}},{{{y=4(1)(-1)=-4}}}.
(0,-4)
{{{y=4(x-0)^2-4}}}
{{{y=4x^2-4}}}
Or multiplying it out.
{{{y=4(x+1)(x-1)}}}
{{{y=4(x^2-1)}}}
{{{y=4x^2-4}}}
(0,-4)