Solve y = x²: Finding Points Where y = 16

Q: Why are there two x-values when y = 16?

Because parabolas are symmetric! The function $ y = x^2 $ creates a U-shape, so any y-value above the vertex has two corresponding x-values - one positive and one negative.

Q: How do I know which point to choose from the options?

Look for the point that matches one of your calculated x-values. Since we found x = 4 and x = -4, check which option shows either $ (4, 16) $ or $ (-4, 16) $.

Q: What if the question asks for all points where y = 16?

Then you'd list both points: $ (4, 16) $ and $ (-4, 16) $. But multiple choice questions usually only include one of them as an option.

Q: Can I check my answer by plugging it back in?

Absolutely! For the point $ (4, 16) $, substitute x = 4 into the original function: $ y = 4^2 = 16 $. If you get y = 16, your point is correct!

Q: Why do some answer choices have negative y-values?

Those are incorrect distractors! Remember that $ y = x^2 $ always produces non-negative y-values because you're squaring x. The function never goes below the x-axis.

Quadratic Functions with Square Root Solutions

Given the function:

$y=x^2$

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

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Set up and solve

00:03 Substitute appropriate values according to the given data, and solve for X

00:16 Extract the root

00:20 When extracting a root there are 2 solutions, positive and negative

00:31 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Given the function:

$y=x^2$

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

Step-by-step solution

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:

$(4,16)$

Final Answer

$(4,16)$

Key Points to Remember

Essential concepts to master this topic

Rule: When $y = x^2$ , substitute the y-value and solve for x
Technique: For $x^2 = 16$ , take square root: $x = \pm 4$
Check: Verify both solutions: $4^2 = 16$ and $(-4)^2 = 16$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting the negative solution when taking square roots
Don't write x = 4 only when solving x² = 16 = missing half the answer! Square roots always give both positive and negative values since both 4² and (-4)² equal 16. Always write x = ±√16 to get both solutions.

Practice Quiz

Test your knowledge with interactive questions

Complete:

The missing value of the function point:

$$ f(x)=x^2 $$

$$ f(?)=16 $$

FAQ

Everything you need to know about this question

Why are there two x-values when y = 16?

Because parabolas are symmetric! The function $y = x^2$ creates a U-shape, so any y-value above the vertex has two corresponding x-values - one positive and one negative.

How do I know which point to choose from the options?

Look for the point that matches one of your calculated x-values. Since we found x = 4 and x = -4, check which option shows either $(4, 16)$ or $(-4, 16)$ .

What if the question asks for all points where y = 16?

Then you'd list both points: $(4, 16)$ and $(-4, 16)$ . But multiple choice questions usually only include one of them as an option.

Can I check my answer by plugging it back in?

Absolutely! For the point $(4, 16)$ , substitute x = 4 into the original function: $y = 4^2 = 16$ . If you get y = 16, your point is correct!

Why do some answer choices have negative y-values?

Those are incorrect distractors! Remember that $y = x^2$ always produces non-negative y-values because you're squaring x. The function never goes below the x-axis.

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