Question 46112
{{{y = x^2 - 2x}}}
a=1 and b=-2
v(-b/2a,f(x)) this is a form used generally to find the vertex
v(1,-1)
y-intercept:
{{{y = 0^2 - 2(0) = 0}}}
(0,0)
x-intercept:
{{{0 = x^2 - 2x}}}
{{{0 = x(x - 2)}}}
x = 0
(0,0)
and
x - 2 = 0
x = 2
(2,0)
{{{graph(600,600,-10,10,-10,10,x^2-2x)}}}