Answer
$(0,-2)$ is on the graph of the equation.
$(1,-2)$ is not on the graph of the equation.
$(2,-2)$ is on the graph of the equation.
Work Step by Step
$x^{2}+xy+y^{2}=4;$ $(0,-2)$ $,$ $(1,-2)$ $,$ $(2,-2)$
Substitute each point into the equation:
$(0,-2)$
$x^{2}+xy+y^{2}=4$
$0^{2}+(0)(-2)+(-2)^{2}=4$
$0+0+4=4$
$4=4$ True
Substituting the point $(0,-2)$ into the equation makes it true. This point is on the graph of the equation.
$(1,-2)$
$x^{2}+xy+y^{2}=4$
$1^{2}+(1)(-2)+(-2)^{2}=4$
$1-2+4=4$
$3\ne4$ False
Substituting the point $(1,-2)$ into the equation makes it false. This point is not on the graph of the equation.
$(2,-2)$
$x^{2}+xy+y^{2}=4$
$2^{2}+(2)(-2)+(-2)^{2}=4$
$4-4+4=4$
$4=4$ True
Substituting the point $(2,-2)$ into the equation makes it true. This point is on the graph of the equation.