Question 653339


since you have {{{y=-x^2}}}, in standard form {{{y=ax^2+bx+c}}} 

The x-value of the vertex can be found using the formula {{{h=-b/2a}}}

since {{{a=-1}}} and {{{b=0}}}, we will have

{{{h=-0/2(-1)}}}

{{{highlight(h=0)}}}

now, use vertex form to find {{{k}}}

{{{y=a(x-h)^2+k}}}...as you can see, when you compare to your equation, {{{k=0}}}

so, vertex will be at ({{{0}}},{{{0}}})

let's see it on a graph:

{{{ graph( 600, 600, -10, 10, -10, 10, -x^2) }}}