Question 358372
The vertex is half way between the 2 roots
{{{f(x) = -x^2 - 2x + 1}}}
Set {{{f(x) = 0}}}
{{{ -x^2 - 2x + 1 = 0}}}
Using the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = -1}}}
{{{b = -2}}}
{{{c = 1}}}
{{{x = (-(-2) +- sqrt( (-2)^2-4*(-1)*1 ))/(2*(-1)) }}}
{{{x = (2 +- sqrt( 4 + 4 ))/ -2) }}}
{{{x = (2 +- sqrt( 8 ))/ -2) }}}
{{{x = (2 +- 2*sqrt( 2 ))/ -2) }}}
{{{x = -1 + sqrt(2)}}}
and
{{{x = -1 - sqrt(2)}}}
{{{x = -1}}} is half way between these 2 roots
Actually, this equals {{{-b/(2a)}}} and
{{{-b/(2a) = -(-2)/(2*(-1))}}}
{{{ -(-2)/(2*(-1)) = -1}}}
Now plug {{{x = -1}}} into equation:
{{{f(x) = -x^2 - 2x + 1}}}
{{{f(-1) = -(-1)^2 - 2*(-1) + 1}}}
{{{f(-1) = -1 + 2 + 1}}}
{{{f(-1) = 2}}}
The vertex is at (-1,2)
Here's a plot:
{{{ graph( 400, 400, -5, 5, -5, 5, -x^2 - 2x + 1) }}}