Solve -3x² + 12x > 0: Finding Positive Function Values

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

Look at the coefficient of x²! If it's negative (like -3), the parabola opens downward. If it's positive, it opens upward. This determines where the function is positive.

Q: What if I get the roots wrong?

Always double-check your factoring! For $ -3x^2 + 12x = 0 $, factor out -3x first: $ -3x(x - 4) = 0 $. Then set each factor to zero.

Q: How can I verify my interval is correct?

Pick a test point inside your interval! Try x = 2: $ -3(2)^2 + 12(2) = -12 + 24 = 12 > 0 $ ✓ Since 12 > 0, our interval is correct.

Q: Do I include the endpoints x = 0 and x = 4?

No! The inequality is $ f(x) > 0 $, not $ f(x) ≥ 0 $. At x = 0 and x = 4, the function equals zero, so we use open intervals: \( 0 .

Quadratic Inequalities with Downward Parabolas

Look at the following function:

$y=-3x^2+12x$

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

$f(x) > 0$

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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:

$y=-3x^2+12x$

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

$f(x) > 0$

Step-by-step solution

To determine where the function $f(x) = -3x^2 + 12x$ is greater than zero, we need to find the roots of the equation.

Step 1: Set the function equal to zero to find the zeros or roots:

$-3x^2 + 12x = 0$

Step 2: Factor the equation:

$-3x(x - 4) = 0$

Setting each factor equal to zero gives us the roots:

$-3x = 0 \rightarrow x = 0$
$x - 4 = 0 \rightarrow x = 4$

Step 3: Since the parabola opens downwards (as the coefficient of $x^2$ is negative), $f(x) > 0$ for $x$ values between the roots. Thus, the function is positive between $x = 0$ and $x = 4$ .

Therefore, the solution is the interval $0 < x < 4$ .

In conclusion, the values of $x$ for which the function is greater than zero are $0 < x < 4$ .

Final Answer

$0 < x < 4$

Key Points to Remember

Essential concepts to master this topic

Rule: Factor out common terms to find zeros first
Technique: Factor $-3x^2 + 12x = -3x(x - 4) = 0$
Check: Test x = 2: $-3(4) + 24 = 12 > 0$ ✓

Common Mistakes

Avoid these frequent errors

Ignoring the negative coefficient of x²
Don't assume the parabola opens upward when solving $-3x^2 + 12x > 0$ = wrong interval! The negative coefficient means the parabola opens downward, so the function is positive BETWEEN the roots. Always check the leading coefficient to determine parabola direction.

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 the parabola opens up or down?

Look at the coefficient of x²! If it's negative (like -3), the parabola opens downward. If it's positive, it opens upward. This determines where the function is positive.

Why is the function positive between x = 0 and x = 4?

Since the parabola opens downward, it's shaped like an upside-down U. The function starts negative, becomes positive between the roots, then negative again. Think of it as the "hump" of the parabola.

What if I get the roots wrong?

Always double-check your factoring! For $-3x^2 + 12x = 0$ , factor out -3x first: $-3x(x - 4) = 0$ . Then set each factor to zero.

How can I verify my interval is correct?

Pick a test point inside your interval! Try x = 2: $-3(2)^2 + 12(2) = -12 + 24 = 12 > 0$ ✓ Since 12 > 0, our interval is correct.

Do I include the endpoints x = 0 and x = 4?

No! The inequality is $f(x) > 0$ , not $f(x) ≥ 0$ . At x = 0 and x = 4, the function equals zero, so we use open intervals: $0 < x < 4$ .

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