Answer
$(1,1)$ is not on the graph of the equation.
$(1,\frac{1}{2})$ is on the graph of the equation.
$(-1,\frac{1}{2})$ is on the graph of the equation.
Work Step by Step
$y(x^{2}+1)=1;$ $(1,1)$ $,$ $(1,\frac{1}{2})$ $,$ $(-1,\frac{1}{2})$
Substitute each point into the equation:
$(1,1)$
$y(x^{2}+1)=1$
$(1)(1^{2}+1)=1$
$(1)(2)=1$
$2\ne1$ False
Substituting the point $(1,1)$ into the equation makes it false. This point is not on the graph of the equation.
$(1,\frac{1}{2})$
$y(x^{2}+1)=1$
$\Big(\dfrac{1}{2}\Big)(1^{2}+1)=1$
$\Big(\dfrac{1}{2}\Big)(2)=1$
$1=1$ True
Substituting the point $(1,\frac{1}{2})$ into the equation makes it true. This point is on the graph of the equation.
$(-1,\frac{1}{2})$
$y(x^{2}+1)=1$
$\Big(\dfrac{1}{2}\Big)[(-1)^{2}+1]=1$
$\Big(\dfrac{1}{2}\Big)(1+1)=1$
$\Big(\dfrac{1}{2}\Big)(2)=1$
$1=1$ True
Substituting the point $(-1,\frac{1}{2})$ into the equation makes it true. This point is on the graph of the equation.