Solve f(x) > 0 on the Graph: Finding Specific x Values

Q: How do I know which intervals to choose from the graph?

Look for where the curve is above the x-axis (positive y-values). The parabola opens upward, so it's positive between the zeros and negative outside them.

Q: Why isn't the answer x > -6 since that's where the parabola goes up?

The vertex at x = -6 is the lowest point, but we need where f(x) > 0. The function is only positive between the two zeros at x = -11 and x = -1.

Q: Do I include the endpoints -11 and -1 in my answer?

No! At x = -11 and x = -1, the function equals zero (f(x) = 0). We want f(x) > 0, so use strict inequality: -11

Q: How can I verify my interval is correct?

Pick any test point inside your interval, like x = -6. If f(-6) > 0 from the graph, your interval is correct. Points outside should give f(x)

Q: What if the parabola opened downward instead?

If the parabola opened downward, it would be positive outside the zeros (x -1) and negative between them. Always check the parabola's direction!

Quadratic Inequalities with Graph Analysis

Find all values of $x$

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

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find all values of $x$

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

Step-by-step solution

To solve the given problem using the graph, we need to determine the intervals along the x-axis where the quadratic function $f(x)$ is positive, based on its x-intercepts $x = -11$ and $x = -1$ as shown on the graph.

Step 1: Identify the x-intercepts from the graph: $x = -11$ and $x = -1$ .
Step 2: Interpret the graph of the quadratic function. Since it is a parabola opening upwards and touches the x-axis at $x = -11$ and $x = -1$ , these are points where the quadratic changes sign.
Step 3: Determine the intervals: The graph is above the x-axis (positive) between the x-intercepts because the parabola is opening upwards. Therefore, the function is positive for $-11 < x < -1$ .

The conclusion is that the quadratic function $f(x)$ is greater than zero in the interval $-11 < x < -1$ .

Therefore, the correct answer is $\mathbf{-11 < x < -1}$ .

Final Answer

$-11 < x < -1$

Key Points to Remember

Essential concepts to master this topic

Sign Analysis: Function is positive where graph sits above x-axis
Technique: Find zeros at x = -11 and x = -1 from graph intercepts
Check: Test x = -6 in middle: f(-6) > 0 confirms interval ✓

Common Mistakes

Avoid these frequent errors

Reading where function equals zero instead of greater than zero
Don't identify the x-intercepts as your answer = gives f(x) = 0, not f(x) > 0! Students confuse solving f(x) = 0 with f(x) > 0. Always look for intervals where the graph is above the x-axis, not touching it.

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 which intervals to choose from the graph?

Look for where the curve is above the x-axis (positive y-values). The parabola opens upward, so it's positive between the zeros and negative outside them.

Why isn't the answer x > -6 since that's where the parabola goes up?

The vertex at x = -6 is the lowest point, but we need where f(x) > 0. The function is only positive between the two zeros at x = -11 and x = -1.

Do I include the endpoints -11 and -1 in my answer?

No! At x = -11 and x = -1, the function equals zero (f(x) = 0). We want f(x) > 0, so use strict inequality: -11 < x < -1.

How can I verify my interval is correct?

Pick any test point inside your interval, like x = -6. If f(-6) > 0 from the graph, your interval is correct. Points outside should give f(x) < 0.

What if the parabola opened downward instead?

If the parabola opened downward, it would be positive outside the zeros (x < -11 or x > -1) and negative between them. Always check the parabola's direction!

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