Question 131282
The equation is in the form {{{y = ax^2 + bx + c}}}
{{{a = -1}}}
{{{b = 2}}}
{{{c = -1}}}
The vertex will be at (-(b/2a), y) where you figure out the
x-coordinate first, then figure out the y-coordinate
{{{-(b/2a) = -(2/(2*(-1)))}}}
{{{-(b/2a) = -2/-2}}}
{{{-(b/2a) = 1}}}
Now plug {{{x = 1}}} into the equation to find the y-coordinate
of the vertex.
{{{y = -x^2 + 2x - 1}}}
{{{y = -(1)^2 + 2*1 - 1}}}
{{{y = -1 + 2 -1}}}
{{{y = 0}}}
So, the vertex is at (1,0), which means it touches the x-axis
at 1 point only
Because {{{a}}} is negative, the parabola opens down, like a cup
spilling out it's water. Therefore it lies below the x-axis
I'll graph it just to make sure
{{{ graph( 600, 600, -10, 10, -10, 3, -x^2 + 2x - 1) }}}