Find the Positive x-values in -x² - 6x - 8 > 0

Q: Why do I need to find where the function equals zero first?

The zeros (roots) are where the parabola crosses the x-axis! These points divide the number line into intervals where the function is either all positive or all negative. Without finding them, you can't determine where $ f(x) > 0 $.

Q: How do I choose good test points for each interval?

Pick any number that's clearly inside each interval. For $ x , try \( x = -5 $. For $ -4 , try \( x = -3 $. For $ x > -2 $, try $ x = 0 $. Simple integers work best!

Q: What if I get the wrong sign when testing points?

Double-check your arithmetic! Remember that $ -x^2 $ means the coefficient of $ x^2 $ is negative. For example: $ f(-3) = -(-3)^2 - 6(-3) - 8 = -9 + 18 - 8 = 1 $.

Q: Why is the answer -4 < x < -2 and not the other intervals?

Because we want $ f(x) > 0 $ (function is positive)! When we tested $ x = -3 $, we got $ f(-3) = 1 $, which is positive. The other intervals gave us negative values.

Q: Does it matter that this is a downward-opening parabola?

Yes! Since the coefficient of $ x^2 $ is negative (-1), this parabola opens downward. This means it's only positive between its roots, not outside them like an upward-opening parabola would be.

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

Look at the following function:

$y=-x^2-6x-8$

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

$f(x) > 0$

2

Step-by-step solution

To determine the values of $x$ for which the quadratic function $y = -x^2 - 6x - 8$ is greater than 0, we will first find the roots of the quadratic equation where it equals zero.

We apply the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substitute $a = -1$ , $b = -6$ , and $c = -8$ into the quadratic formula:

x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-1)(-8)}}{2(-1)}

Simplifying inside the square root and the rest of the expression:

x = \frac{6 \pm \sqrt{36 - 32}}{-2}

x = \frac{6 \pm \sqrt{4}}{-2}

Since $\sqrt{4} = 2$ , the equation becomes:

x = \frac{6 \pm 2}{-2}

This gives us two potential solutions:

- $x = \frac{8}{-2} = -4$

- $x = \frac{4}{-2} = -2$

The roots divide the x-axis into three intervals: $x < -4$ , $-4 < x < -2$ , and $x > -2$ .

To find where the function is positive, choose test points from these intervals:

For $x < -4$ (e.g., $x = -5$ ): $f(-5) = -(-5)^2 - 6(-5) - 8 = -25 + 30 - 8 = -3$
For $-4 < x < -2$ (e.g., $x = -3$ ): $f(-3) = -(-3)^2 - 6(-3) - 8 = -9 + 18 - 8 = 1$
For $x > -2$ (e.g., $x = 0$ ): $f(0) = -(0)^2 - 6(0) - 8 = -8$

From this, the function is positive on the interval $-4 < x < -2$ .

Therefore, the solution to the problem is $-4 < x < -2$ .

3

Final Answer

$-4 < x < -2$

Key Points to Remember

Essential concepts to master this topic

Find Zeros First: Set equation equal to zero and solve using quadratic formula
Test Points: Pick values from each interval: f(-5) = -3, f(-3) = 1, f(0) = -8
Check Intervals: Function is positive only where test point gives positive value ✓

Common Mistakes

Avoid these frequent errors

Forgetting to test intervals between roots
Don't just find the roots x = -4 and x = -2 and guess the answer! This leads to picking the wrong interval or even the opposite solution. Always test a point from each interval created by the roots to see where the function is actually positive or negative.

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

Why do I need to find where the function equals zero first?

+

The zeros (roots) are where the parabola crosses the x-axis! These points divide the number line into intervals where the function is either all positive or all negative. Without finding them, you can't determine where $f(x) > 0$ .

How do I choose good test points for each interval?

+

Pick any number that's clearly inside each interval. For $x < -4$ , try $x = -5$ . For $-4 < x < -2$ , try $x = -3$ . For $x > -2$ , try $x = 0$ . Simple integers work best!

What if I get the wrong sign when testing points?

+

Double-check your arithmetic! Remember that $-x^2$ means the coefficient of $x^2$ is negative. For example: $f(-3) = -(-3)^2 - 6(-3) - 8 = -9 + 18 - 8 = 1$ .

Why is the answer -4 < x < -2 and not the other intervals?

+

Because we want $f(x) > 0$ (function is positive)! When we tested $x = -3$ , we got $f(-3) = 1$ , which is positive. The other intervals gave us negative values.

Does it matter that this is a downward-opening parabola?

+

Yes! Since the coefficient of $x^2$ is negative (-1), this parabola opens downward. This means it's only positive between its roots, not outside them like an upward-opening parabola would be.

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Find the Positive x-values in -x² - 6x - 8 > 0

Quadratic Inequalities with Test Point Method

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Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why do I need to find where the function equals zero first?

How do I choose good test points for each interval?

What if I get the wrong sign when testing points?

Why is the answer -4 < x < -2 and not the other intervals?

Does it matter that this is a downward-opening parabola?

🌟 Unlock Your Math Potential

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