Answer:
The ordered pairs are(missing values in original question are underlined)
(2, 2), (3, 7), (-4,14),(4,14), (-1,-1),(1,-1)
Explanation:
The given function is
![y=x^(2)-2](https://img.qammunity.org/2023/formulas/mathematics/high-school/p4m1m9xje6nt4yqyi7q2ar9ftgihene3w8.png)
In the first two ordered pairs we are given the value of
and all we have to do is substitute for
in the above function and solve for
![y](https://img.qammunity.org/2023/formulas/mathematics/high-school/39evgwyfztrxf0jqm5m4q20wdsbwaeh9qb.png)
So, for (2,...) where
![x=2,](https://img.qammunity.org/2023/formulas/mathematics/high-school/bantus9fqtp21oquv6poto8jtsdrcky56b.png)
![y=2^(2)2-2=4-2=\boldsymbol{2}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ehaf4icliu8q4suqx7vyk8o6r9k8srg3gj.png)
For the second pair, (3,...)
![y=3^(2)-2=9-2=\boldsymbol{7}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ifhu88azdspaly3wtemrqll5zbjgk8mtf1.png)
For the last two pairs, we are given what the y value is and we substitute for y in the function equation and solve for x
For (..., 14) we get
14=x^{2}-2==>16=x^{2},so x=\pm4. This means the two possible values for x are -4 and 4
For the last pair, y=-1 which gives us
which gives the two possible values for x as
and
![x=\bold{1}](https://img.qammunity.org/2023/formulas/mathematics/high-school/iz2mbwwgimwqqwyw4w4o85pdmr8xq7hx8g.png)
See the attached image for a visual depiction of these points