Determine X Values for Positive Outputs in y=-x²-5x

Q: Why do I need to test points in each interval?

Because quadratic functions change signs at their zeros! The parabola goes from positive to negative (or vice versa) as you cross each zero, so testing confirms which intervals are actually positive.

Q: How do I choose good test points?

Pick simple integers that fall clearly inside each interval. For interval (-5, 0), choose x = -1 or x = -2. Avoid the zeros themselves since we want f(x) > 0, not f(x) ≥ 0.

Q: Why isn't the answer the whole parabola above the x-axis?

This parabola opens downward (coefficient of x² is negative), so it's only positive between its zeros. If it opened upward, the positive region would be outside the zeros!

Q: Do the endpoints x = -5 and x = 0 count?

No! The inequality is $ f(x) > 0 $ (strictly greater than), so points where f(x) = 0 are excluded. Use open interval notation: (-5, 0).

Quadratic Inequalities with Sign Analysis

Look at the following function:

$y=-x^2-5x$

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:

$y=-x^2-5x$

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

$f(x) > 0$

Step-by-step solution

To determine for which values of $x$ the quadratic function is positive, follow these steps:

Step 1: Set the quadratic equation to zero: $-x^2 - 5x = 0$ .
Step 2: Factor the equation: $-x(x + 5) = 0$ .
Step 3: Solve for the roots: $x = 0$ and $x = -5$ .
Step 4: Determine the intervals defined by the roots: $(-\infty, -5)$ , $(-5, 0)$ , and $(0, \infty)$ .
Step 5: Test each interval:

For interval $(-\infty, -5)$ , choose $x = -6$ : $-(-6)^2 - 5(-6) = -36 + 30 = -6$ (negative).
For interval $(-5, 0)$ , choose $x = -1$ : $-(-1)^2 - 5(-1) = -1 + 5 = 4$ (positive).
For interval $(0, \infty)$ , choose $x = 1$ : $-(1)^2 - 5(1) = -1 - 5 = -6$ (negative).

Step 6: The function is positive in the interval $(-5, 0)$ .

Thus, the solution to the inequality $f(x) > 0$ is $-5 < x < 0$ .

Final Answer

$-5 < x < 0$

Key Points to Remember

Essential concepts to master this topic

Zeros: Find where quadratic equals zero by factoring
Intervals: Test points between zeros: x = -1 gives 4 (positive)
Check: Verify interval endpoints are excluded from solution ✓

Common Mistakes

Avoid these frequent errors

Testing only one interval or guessing signs
Don't assume the parabola is positive everywhere between zeros = wrong solution! The sign changes at each zero, so you might get the opposite interval. Always test a point in each interval created by the zeros.

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

Why do I need to test points in each interval?

Because quadratic functions change signs at their zeros! The parabola goes from positive to negative (or vice versa) as you cross each zero, so testing confirms which intervals are actually positive.

How do I choose good test points?

Pick simple integers that fall clearly inside each interval. For interval (-5, 0), choose x = -1 or x = -2. Avoid the zeros themselves since we want f(x) > 0, not f(x) ≥ 0.

What if I get the wrong interval?

Double-check your factoring and test point calculations! A common error is sign mistakes when substituting negative values. Work step-by-step: $-(-1)^2 - 5(-1) = -1 + 5 = 4$ .

Why isn't the answer the whole parabola above the x-axis?

This parabola opens downward (coefficient of x² is negative), so it's only positive between its zeros. If it opened upward, the positive region would be outside the zeros!

Do the endpoints x = -5 and x = 0 count?

No! The inequality is $f(x) > 0$ (strictly greater than), so points where f(x) = 0 are excluded. Use open interval notation: (-5, 0).

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