Question 983839
{{{y=(x+1)(x-2)}}}


You can make a table of a few x y values; and get the zeros from the factors of y.
Zeros for y are x=-1 and x=2.  The vertex occurs at  {{{(-1+2)/2=1/2}}}; at which {{{y=(1/2+1)(1/2-2)=(3/2)(-3/2)=-9/4}}}  which is also {{{-2&1/2}}}.  Vertex  ( 1/2, -9/4 ).


{{{graph(300,300,-3,5,-4,4,x^2-x-2)}}}