Solve the Quadratic Inequality: Finding X When -x²-4 > 0

Q: How do I know if a parabola opens upward or downward?

Look at the coefficient of $ x^2 $! If it's positive, the parabola opens upward (U-shape). If it's negative (like -1 in our problem), it opens downward (∩-shape).

Q: What's the vertex and why is it important?

The vertex is the turning point of a parabola. For downward parabolas like $ y = -x^2 - 4 $, it's the highest point. If this highest point is below the x-axis, the entire parabola is negative!

Q: Can I solve this by setting the function equal to zero?

That would find where the parabola crosses the x-axis, but we need where it's above the x-axis. Since our parabola never crosses (discriminant

Q: How do I find the vertex of any quadratic?

For $ y = ax^2 + bx + c $, the vertex x-coordinate is $ x = -\frac{b}{2a} $. In our case: a = -1, b = 0, so $ x = -\frac{0}{2(-1)} = 0 $.

Q: Why is the answer 'No x' and not 'All x'?

We need $ f(x) > 0 $ (positive values). Since the maximum value is -4 and the parabola only goes down from there, every y-value is negative. No x-values make the function positive!

Quadratic Inequalities with Negative Leading Coefficients

Given the function:

$y=-x^2-4$

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:

$y=-x^2-4$

Determine for which values of x the following holds:

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

Step-by-step solution

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

Step 1: Identify the form and behavior of the given quadratic function.
Step 2: Analyze the vertical position of the function's vertex.
Step 3: Determine inequality feasibility for positive $y$ .

Step 1: The function given is $y = -x^2 - 4$ , which is a parabola opening downwards because of the negative coefficient of $x^2$ .

Step 2: The vertex of this parabola occurs at $x = -\frac{b}{2a} = 0$ , considering no $x$ term present (i.e., $b = 0$ ). The vertex value of $y$ is $f(0) = -0^2 - 4 = -4$ .

Step 3: The vertex, being the highest point at $(0, -4)$ , means the entire parabola remains below the x-axis.

Since the maximum point is negative and the parabola only descends from this vertex, the function $y = -x^2 - 4$ remains less than zero for every $x$ . There are no $x$ values making $y$ positive.

Therefore, the solution is No x.

Final Answer

No x

Key Points to Remember

Essential concepts to master this topic

Vertex Analysis: Find maximum point for downward parabolas using vertex formula
Technique: Calculate vertex at x = 0 gives y = -4
Check: Since maximum value is -4, entire parabola stays below x-axis ✓

Common Mistakes

Avoid these frequent errors

Assuming parabolas can be positive somewhere
Don't think a downward parabola with negative vertex must cross x-axis = wrong assumption! When the maximum value is negative, the entire function stays negative. Always check the vertex value to determine if positive solutions exist.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below does not intersect the $$ x $$ -axis.

The parabola's vertex is marked A.

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

FAQ

Everything you need to know about this question

How do I know if a parabola opens upward or downward?

Look at the coefficient of $x^2$ ! If it's positive, the parabola opens upward (U-shape). If it's negative (like -1 in our problem), it opens downward (∩-shape).

What's the vertex and why is it important?

The vertex is the turning point of a parabola. For downward parabolas like $y = -x^2 - 4$ , it's the highest point. If this highest point is below the x-axis, the entire parabola is negative!

Can I solve this by setting the function equal to zero?

That would find where the parabola crosses the x-axis, but we need where it's above the x-axis. Since our parabola never crosses (discriminant < 0), there are no positive values.

How do I find the vertex of any quadratic?

For $y = ax^2 + bx + c$ , the vertex x-coordinate is $x = -\frac{b}{2a}$ . In our case: a = -1, b = 0, so $x = -\frac{0}{2(-1)} = 0$ .

Why is the answer 'No x' and not 'All x'?

We need $f(x) > 0$ (positive values). Since the maximum value is -4 and the parabola only goes down from there, every y-value is negative. No x-values make the function positive!

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