Determine X Values for y = -x² - 4 Less than Zero

Q: Why doesn't this quadratic ever equal zero?

Because $ y = -x^2 - 4 $ has its maximum value at the vertex (0, -4). Since the parabola opens downward and the highest point is -4, the function is always less than zero!

Q: How do I know the parabola opens downward?

Look at the coefficient of x²! When it's negative (like the -1 in $ -x^2 $), the parabola opens downward. When positive, it opens upward.

Q: What if the question asked when f(x) > 0 instead?

Then the answer would be "No x" because this function is never positive. It's always negative or at most -4 at its highest point.

Q: How do I find the vertex of this parabola?

For $ y = -x^2 - 4 $, the vertex is at x = 0 (since there's no x-term), and y = -4. So the vertex is (0, -4).

Q: Can I graph this to check my answer?

Absolutely! Graph $ y = -x^2 - 4 $ and you'll see the entire parabola sits below the x-axis. This visually confirms that f(x)

Quadratic Inequalities with Always-Negative Functions

Given the function:

$y=-x^2-4$

Determine for which values of x the following is true:

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

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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 is true:

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

Step-by-step solution

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

Step 1: Analyze the quadratic function given: $y = -x^2 - 4$ .
Step 2: Analyze what $y < 0$ implies for the values of $x$ .

Now, let's solve it:

Step 1: Consider the given quadratic function $y = -x^2 - 4$ . The function $y = -x^2$ represents a downward-opening parabola due to the negative sign before $x^2$ . The entire graph of the parabola, being shifted downward by 4 units with the term $-4$ , will be wholly beneath the x-axis since there's no positive vertex or value. The vertex of the parabola is at $(0, -4)$ , which is already below zero.

Step 2: Because the quadratic term causes the graph to be a parabola opening downwards, it means that for any $x$ , $y = -x^2 - 4 \leq -4$ , which is always less than zero. Thus, the inequality $f(x) = -x^2 - 4 < 0$ is satisfied for all real numbers $x$ .

Therefore, the solution is that the inequality is true for all $x$ .

Final Answer

All x

Key Points to Remember

Essential concepts to master this topic

Rule: Quadratic functions with negative leading coefficient and negative vertex are always negative
Technique: For $y = -x^2 - 4$ , vertex at (0, -4) confirms maximum value is -4
Check: Test any x-value: when x = 0, y = -4 < 0 ✓

Common Mistakes

Avoid these frequent errors

Trying to solve by setting the function equal to zero
Don't solve -x² - 4 = 0 to find boundary points = no real solutions exist! This function never equals zero since it's always negative. Always analyze the parabola's position relative to the x-axis using the vertex and direction.

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 doesn't this quadratic ever equal zero?

Because $y = -x^2 - 4$ has its maximum value at the vertex (0, -4). Since the parabola opens downward and the highest point is -4, the function is always less than zero!

How do I know the parabola opens downward?

Look at the coefficient of x²! When it's negative (like the -1 in $-x^2$ ), the parabola opens downward. When positive, it opens upward.

What if the question asked when f(x) > 0 instead?

Then the answer would be "No x" because this function is never positive. It's always negative or at most -4 at its highest point.

How do I find the vertex of this parabola?

For $y = -x^2 - 4$ , the vertex is at x = 0 (since there's no x-term), and y = -4. So the vertex is (0, -4).

Can I graph this to check my answer?

Absolutely! Graph $y = -x^2 - 4$ and you'll see the entire parabola sits below the x-axis. This visually confirms that f(x) < 0 for all x-values.

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