Solve the Quadratic Function Inequality: When is y = -0.9x² Less Than Zero?

Q: How can I visualize this problem?

Picture an upside-down parabola with its vertex touching the x-axis at (0,0). The function is negative everywhere except at that single point where it touches zero.

Q: What if the coefficient was positive instead of negative?

If we had $ y = 0.9x^2 $, the parabola would open upward and would never be negative! It would be zero at x = 0 and positive everywhere else.

Q: Why do we write the answer as x ≠ 0 instead of listing ranges?

Since the function is negative for all real numbers except zero, writing $ x \neq 0 $ is the most concise and complete way to express this solution.

Q: How do I check my answer is correct?

Test a few values! Try x = 1: $ -0.9(1)^2 = -0.9 ✓ Try x = -2: \( -0.9(-2)^2 = -3.6 ✓ Try x = 0: \( -0.9(0)^2 = 0 $ (not

Quadratic Inequality Analysis with Zero Boundaries

Given the function:

$y=-0.9x^2$

Determine for which values of x the following holds:

$f\left(x\right) < 0$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Given the function:

$y=-0.9x^2$

Determine for which values of x the following holds:

$f\left(x\right) < 0$

Step-by-step solution

To solve this problem, we need to determine when the quadratic function $y = -0.9x^2$ is less than zero. This requires analyzing the entire set of x-values.

Given that the function $y = -0.9x^2$ is a parabola opening downward because $a = -0.9 < 0$ , the function will take negative values for all $x$ except when $x^2 = 0$ .
The vertex of this parabola is at the origin, $x = 0$ , since terms for linear ( $bx$ ) and constant ( $c$ ) are zero in this function.
At $x = 0$ , $y = -0.9(0)^2 = 0$ , meaning at $x = 0$ , the function is exactly zero.
For any $x \neq 0$ , the term $x^2 > 0$ , hence $-0.9x^2 < 0$ , indicating the function is negative.

Therefore, the function $y = -0.9x^2$ is less than zero for all values except at $x = 0$ .

Consequently, the solution to the problem is that the function is negative for all $x \ne 0$ .

This corresponds to choice: $x\ne0$ .

Final Answer

$x\ne0$

Key Points to Remember

Essential concepts to master this topic

Parabola Direction: Coefficient -0.9 < 0 means parabola opens downward
Critical Point: At x = 0, function equals zero: -0.9(0)² = 0
Verify: Test any non-zero value like x = 1: -0.9(1)² = -0.9 < 0 ✓

Common Mistakes

Avoid these frequent errors

Forgetting that x = 0 makes the function equal zero, not negative
Don't include x = 0 in your solution set = wrong answer! At x = 0, the function equals zero (-0.9 × 0² = 0), which doesn't satisfy f(x) < 0. Always check what happens at critical points like x = 0.

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 $f\left(x\right) > 0$ .

FAQ

Everything you need to know about this question

Why isn't x = 0 part of the solution if the parabola opens downward?

Great question! Even though the parabola opens downward, at x = 0 the function equals exactly zero, not negative. We need $f(x) < 0$ , so zero doesn't count.

How can I visualize this problem?

Picture an upside-down parabola with its vertex touching the x-axis at (0,0). The function is negative everywhere except at that single point where it touches zero.

What if the coefficient was positive instead of negative?

If we had $y = 0.9x^2$ , the parabola would open upward and would never be negative! It would be zero at x = 0 and positive everywhere else.

Why do we write the answer as x ≠ 0 instead of listing ranges?

Since the function is negative for all real numbers except zero, writing $x \neq 0$ is the most concise and complete way to express this solution.

How do I check my answer is correct?

Test a few values! Try x = 1: $-0.9(1)^2 = -0.9 < 0$ ✓ Try x = -2: $-0.9(-2)^2 = -3.6 < 0$ ✓ Try x = 0: $-0.9(0)^2 = 0$ (not < 0)

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