Solve the Quadratic Inequality: When Does y = -4x^2 + 24x Become Positive?

Q: How do I know if the parabola opens up or down?

Look at the leading coefficient (the number in front of $ x^2 $). If it's positive, the parabola opens upward. If it's negative (like -4 here), it opens downward.

Q: Why is the function positive between the roots?

Since our parabola opens downward, it starts negative, becomes positive between the roots, then goes negative again. Think of it like an upside-down U shape!

Q: Do I include the endpoints 0 and 6 in my answer?

No! The inequality is $ f(x) > 0 $ (greater than), not $ f(x) ≥ 0 $. At $ x = 0 $ and $ x = 6 $, the function equals zero, not greater than zero.

Q: How can I double-check my interval is correct?

Pick any number inside your interval and substitute it into the original function. For example, try $ x = 3 $: $ -4(3)^2 + 24(3) = -36 + 72 = 36 > 0 $ ✓

Quadratic Inequalities with Negative Leading Coefficients

Look at the following function:

$y=-4x^2+24x$

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 following function:

$y=-4x^2+24x$

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

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

Step-by-step solution

To determine the values of $x$ for which the function $y = -4x^2 + 24x$ is greater than zero, we will proceed as follows:

Step 1: Find the roots of the quadratic equation.

We start by solving $-4x^2 + 24x = 0$ to find the critical points. This can be factored as:

$x(-4x + 24) = 0$

This equation gives us two roots:

$x = 0$
$-4x + 24 = 0 \Rightarrow x = 6$

Step 2: Determine intervals for positivity.

The roots divide the number line into three intervals: $(-\infty, 0)$ , $(0, 6)$ , and $(6, \infty)$ .

Since the parabola opens downwards (as indicated by the negative leading coefficient), the function will be positive between the roots:

$0 < x < 6$

Conclusion: To ensure the function $y = -4x^2 + 24x$ is greater than zero, the value of $x$ must be between 0 and 6.

Therefore, the solution to the problem is $0 < x < 6$ .

Final Answer

$0 < x < 6$

Key Points to Remember

Essential concepts to master this topic

Rule: Factor quadratic and find roots to identify critical points
Technique: Test intervals: for $x = 3$ , get $-4(9) + 24(3) = 36 > 0$
Check: Verify endpoints: at $x = 0$ and $x = 6$ , function equals zero ✓

Common Mistakes

Avoid these frequent errors

Confusing parabola direction with inequality solution
Don't assume downward parabolas are always negative! A downward parabola $y = -4x^2 + 24x$ is actually positive between its roots. This confusion leads to choosing outside intervals instead of between roots. Always check the parabola's behavior by testing values in each interval.

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 if the parabola opens up or down?

Look at the leading coefficient (the number in front of $x^2$ ). If it's positive, the parabola opens upward. If it's negative (like -4 here), it opens downward.

Why is the function positive between the roots?

Since our parabola opens downward, it starts negative, becomes positive between the roots, then goes negative again. Think of it like an upside-down U shape!

What if the roots were the same number?

If both roots are equal, the parabola just touches the x-axis at one point. For $f(x) > 0$ , there would be no solution since the function never goes above the x-axis.

Do I include the endpoints 0 and 6 in my answer?

No! The inequality is $f(x) > 0$ (greater than), not $f(x) ≥ 0$ . At $x = 0$ and $x = 6$ , the function equals zero, not greater than zero.

How can I double-check my interval is correct?

Pick any number inside your interval and substitute it into the original function. For example, try $x = 3$ : $-4(3)^2 + 24(3) = -36 + 72 = 36 > 0$ ✓

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