Solving y = x²: Does a Point Exist Where y = -2?

Quadratic Functions with Range Analysis

Given the function:

$y=x^2$

Is there a point for ? $y=-2$ ?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Does the point exist?

00:03 Let's substitute appropriate values according to the given data, and solve to find the point

00:07 Any number squared will always equal a positive number

00:11 Therefore, the point doesn't exist

00:17 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

Given the function:

$y=x^2$

Is there a point for ? $y=-2$ ?

2

Step-by-step solution

To solve this problem, we'll follow these steps:

Step 1: Analyze the equation $y = x^2$ .
Step 2: Investigate whether a negative $y$ -value is possible.

Now, let's work through each step:
Step 1: The function we have is $y = x^2$ . This function is defined for all real numbers and always gives a non-negative value $y$ because squaring a real number cannot result in a negative number.

Step 2: We need to check whether $y = -2$ is possible by solving $x^2 = -2$ . In the real number system, no real number $x$ satisfies this equation since the square of any real number is non-negative.
Therefore, there is no real point where $y = -2$ on the graph of the function $y = x^2$ .

Therefore, the solution to the problem is No.

3

Final Answer

No

Key Points to Remember

Essential concepts to master this topic

Range Property: Squared values are always non-negative for real numbers
Technique: Set $x^2 = -2$ and check if solution exists
Check: Verify minimum value occurs at vertex (0,0) where y ≥ 0 ✓

Common Mistakes

Avoid these frequent errors

Assuming negative y-values are possible for quadratic functions
Don't think that $y = x^2$ can equal -2 just because you can write the equation! Since any real number squared is non-negative, $x^2 \geq 0$ always. Always remember that parabolas opening upward have a minimum y-value at their vertex.

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 can't x² ever equal a negative number?

+

When you square any real number, the result is always positive or zero. For example: $3^2 = 9$ , $(-3)^2 = 9$ , and $0^2 = 0$ . There's no real number that gives a negative when squared!

What's the range of y = x²?

+

The range is all y-values ≥ 0. The parabola has its lowest point at (0,0) and goes upward forever. So y can be 0, 1, 4, 100, but never -1, -2, or any negative value.

Could there be complex solutions to x² = -2?

+

Yes! In the complex number system, $x = ±i\sqrt{2}$ where i is the imaginary unit. But for real-valued functions like $y = x^2$ , we only consider real solutions.

How do I find the minimum value of a parabola?

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For $y = x^2$ , the vertex is at (0,0), which gives the minimum y-value of 0. The parabola opens upward, so this is the lowest point on the graph.

What if the question asked about y = 2 instead?

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Then we'd solve $x^2 = 2$ , giving $x = ±\sqrt{2}$ . Since 2 is positive, real solutions exist! The points would be $(\sqrt{2}, 2)$ and $(-\sqrt{2}, 2)$ .

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