Solve y = x²: Finding Points Where y = 16

Given the function:

$y=x^2$

Is there a point for ? $y=16$?

The problem asks us to find an $x$ such that in the function $y = x^2$, the value of $y$ becomes 16. To do this, we'll substitute $y = 16$ into the equation and solve for $x$.

1. Start with the equation of the function:

$y = x^2$

2. Substitute $y = 16$ into the equation:

$16 = x^2$

3. Solve $x^2 = 16$ for $x$:

• Take the square root of both sides to solve for $x$:
• $x = \pm \sqrt{16}$
• This gives $x = 4$ or $x = -4$

4. Identify the points on the function for these values of $x$:

• For $x = 4$, the point is $(4, 16)$.
• For $x = -4$, the point is $(-4, 16)$, but this is not provided in the choice list.

Among the given options, the point we find in the choices is:

$(4, 16)$

Therefore, the correct answer is the choice that corresponds with this point:

$(4,16)$