Determine X Values where the Quadratic Equation y = -x² is Positive

Given the function:

y=x2 y=-x^2

Determine for which values of x is f(x)>0 f\left(x\right) > 0 true

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1

Understand the problem

Given the function:

y=x2 y=-x^2

Determine for which values of x is f(x)>0 f\left(x\right) > 0 true

2

Step-by-step solution

To solve the problem, we need to understand when the function y=x2 y = -x^2 is greater than 0.

1. **Analysis of the Function**: The given function is y=x2 y = -x^2 . Here, y y represents the output of the quadratic expression, where the coefficient of x2 x^2 is negative (1-1). This means that for every input x x , the output is the negative of x2 x^2 .

2. **Properties of the Square**: The expression x2 x^2 is always non-negative, i.e., x20 x^2 \geq 0 for all real numbers x x . This implies:

  • For x=0 x = 0 , x2=0 x^2 = 0 .
  • For any non-zero x x , x2>0 x^2 > 0 .

3. **Impact of the Negative Sign**: Since y=x2 y = -x^2 :

  • If x2=0 x^2 = 0 (which happens when x=0 x = 0 ), then y=0 y = 0 .
  • If x2>0 x^2 > 0 (which happens when x0 x \ne 0 ), then y<0 y < 0 .

4. **Conclusion on Positivity**: There are no values of x x for which y=x2 y = -x^2 is greater than 0, as y0 y \leq 0 for all real x x . Thus, the function is never positive.

Therefore, the solution to this problem is No x x .

3

Final Answer

x0 x\ne0

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of \( x \) where \( f\left(x\right) > 0 \).

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