Solve the Quadratic Inequality: When is -4x² - 12 Positive?

Q: How do I know the parabola opens downward?

Look at the coefficient of $ x^2 $! Since it's negative (-4), the parabola opens downward like an upside-down U. Positive coefficients make parabolas open upward.

Q: Why don't I need to solve -4x² - 12 = 0?

We want $ f(x) > 0 $, not $ f(x) = 0 $! Since this parabola's highest point is at y = -12 (below zero), it never becomes positive.

Q: What if the question asked for f(x) < 0 instead?

Then the answer would be all real numbers! Since this parabola is always negative (maximum is -12), every x-value makes f(x)

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

Use $ x = -\frac{b}{2a} $ for the x-coordinate, then substitute back to find y. For $ y = -4x^2 - 12 $, we have a = -4, b = 0, so x = 0.

Q: Could this parabola ever have a different answer?

No! The structure $ y = -4x^2 - 12 $ means we start at -12 and subtract more (since $ -4x^2 \leq 0 $). This function is always negative.

Quadratic Inequalities with Downward Parabolas

Look at the function below:

$y=-4x^2-12$

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

Look at the function below:

$y=-4x^2-12$

Determine for which values of $x$ the following is true:

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

Step-by-step solution

The goal is to find the values of $x$ for which $f(x) > 0$ given $y = -4x^2 - 12$ . Start by analyzing the equation $y = -4x^2 - 12$ .

Since the quadratic term is negative ( $a = -4$ ), the parabola opens downwards. This means the maximum point of the parabola (its vertex) is at the top.

Find the vertex using the formula for the $x$ -coordinate of the vertex, $x = -\frac{b}{2a}$ . Here, $b = 0$ , so the vertex is at $x = 0$ .

Calculate $y$ -value at the vertex:
$y = -4(0)^2 - 12 = -12$ .

This evaluation confirms $y = -12$ , which is less than 0 at the vertex.

Since the entire parabola opens downward and the highest point $y$ achieves is still negative ( $-12$ ), the function is never greater than 0 at any point.

No $x$ values satisfy $y > 0$ . Therefore, no values of $x$ make the quadratic positive.

Thus, the answer is: No values of $x$ .

Final Answer

No values of $x$

Key Points to Remember

Essential concepts to master this topic

Key Property: Downward parabola with negative coefficient never reaches positive values
Vertex Analysis: Find maximum at x = 0: y = -4(0)² - 12 = -12
Verification: Since maximum y-value is -12 < 0, no x makes f(x) > 0 ✓

Common Mistakes

Avoid these frequent errors

Assuming the parabola crosses the x-axis
Don't set -4x² - 12 = 0 and solve for x-intercepts to find where it's positive! This parabola never touches the x-axis since its maximum is -12. Always check the vertex first to see if the parabola can reach positive values.

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

How do I know the parabola opens downward?

Look at the coefficient of $x^2$ ! Since it's negative (-4), the parabola opens downward like an upside-down U. Positive coefficients make parabolas open upward.

Why don't I need to solve -4x² - 12 = 0?

We want $f(x) > 0$ , not $f(x) = 0$ ! Since this parabola's highest point is at y = -12 (below zero), it never becomes positive.

What if the question asked for f(x) < 0 instead?

Then the answer would be all real numbers! Since this parabola is always negative (maximum is -12), every x-value makes f(x) < 0 true.

How do I find the vertex of any parabola?

Use $x = -\frac{b}{2a}$ for the x-coordinate, then substitute back to find y. For $y = -4x^2 - 12$ , we have a = -4, b = 0, so x = 0.

Could this parabola ever have a different answer?

No! The structure $y = -4x^2 - 12$ means we start at -12 and subtract more (since $-4x^2 \leq 0$ ). This function is always negative.

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