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 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 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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