Question 65282
find the intercepts and vertex for {{{ y=-x^2-4x }}}

When x=0,  y=0.
When y=0, {{{ -x^2-4x = 0 }}}
{{{ x^2+4x = 0 }}}
{{{ x(x+4) = 0 }}}
x = 0 or x = -4.
Hence the intercept points are (0,0) and (-4,0)

{{{ y=-x^2-4x }}}
{{{ -y = x^2 + 4x }}}
{{{ -y + 4 = x^2 + 4x + 4 }}}
{{{ -(y-4) = (x+2)^2 }}}
Hence the vertex is (-2,4)



find intercept for x^3-2x^2-9x=18=0
{{{ x^3-2x^2-9x+18 = 0 }}}
{{{ x^2(x-2)-9(x-2) = 0 }}}
{{{ (x-2)(x^2-9) = 0 }}}
{{{ (x-2)(x-3)(x+3) = 0 }}}
x = 2 , 3 or -3
Hence the intercepts are (2,0) , (3,0) and (-3,0)