Question 1075198
When {{{x=0}}},
{{{y=0^2-0-2}}}
{{{y=-2}}}
(0,-2)
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When {{{y=0}}}
{{{x^2-x-2=0}}}
{{{(x-2)(x+1)=0}}}
Two solutions:
{{{x-2=0}}}
{{{x=2}}}
and
{{{x+1=0}}}
{{{x=-1}}}
(2,0) and (-1,0)
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*[illustration r34.JPG].