Hello :
<span>a reflection over the y-axis</span>
A good place to start is to set to y. That would mean we are looking for to be an integer. Clearly, , because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since is a radical, it only outputs values from , which means y is on the closed interval: .
With that, we don't really have to consider y anymore, since we know the interval that is on.
Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval , which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:
Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.
I think the correct answer is C. because each if forming a right triangle when you trace it.
If i'm correct brainliest please