Parabola Property: When Downward-Facing Vertex Above X-axis Implies Positive Values

Q: If the vertex is above the x-axis, why isn't the whole parabola positive?

The vertex is just the highest point of a downward parabola. As you move away from the vertex, the parabola goes down and can cross below the x-axis, creating negative y-values.

Q: How can I tell if a downward parabola crosses the x-axis?

Use the discriminant $ b^2 - 4ac $. If it's positive, the parabola has two real roots and crosses the x-axis. If it's zero, it touches once. If negative, it never crosses.

Q: What's a simple example of this situation?

Consider $ y = -x^2 + 1 $. The vertex is at $ (0, 1) $ above the x-axis, but when $ x = 2 $, we get $ y = -4 + 1 = -3 $, which is negative!

Q: Does this apply to upward-opening parabolas too?

No! If an upward-opening parabola has its vertex above the x-axis, then the entire parabola is above the x-axis since that's the lowest point.

Q: How far from the vertex do I need to check?

It depends on the parabola! For $ y = -x^2 + c $, the roots are at $ x = ±\sqrt{c} $. Test points beyond these roots to see negative values.

Parabola Analysis with Downward Opening

If a parabola is bending downwards and its vertex is above the x-axis, then it is always positive.

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

If a parabola is bending downwards and its vertex is above the x-axis, then it is always positive.

Step-by-step solution

To solve this problem, we need to carefully examine the nature of the parabolic function given its attributes:

The parabola opens downward, which indicates the coefficient $a < 0$ in its standard form $y = ax^2 + bx + c$ .
The vertex is above the x-axis, suggesting that its y-coordinate, given by $c - \frac{b^2}{4a}$ , is positive.

However, having the vertex above the x-axis alone does not guarantee that the entire parabola remains above the x-axis. A downward-opening parabola can have parts below the x-axis even if its vertex is above it.

Consider a simple example. Take $y = -x^2 + 6$ , which is a downward-opening parabola with vertex $(0, 6)$ . While the vertex is above the x-axis, the roots $x = -\sqrt{6}$ and $x = \sqrt{6}$ indicate that it crosses the x-axis, thus having negative values for $x$ in $(-\sqrt{6}, \sqrt{6})$ .

Therefore, the statement that the parabola is always positive is Incorrect.

Final Answer

Incorrect

Key Points to Remember

Essential concepts to master this topic

Rule: Downward parabolas can cross x-axis despite vertex above it
Technique: Check discriminant: $b^2 - 4ac > 0$ means real roots exist
Check: Test values beyond vertex: $y = -x^2 + 6$ at $x = 3$ gives $y = -3$ ✓

Common Mistakes

Avoid these frequent errors

Assuming vertex position determines entire function sign
Don't think vertex above x-axis means always positive = missing negative portions! This ignores that downward parabolas extend infinitely down. Always check if the parabola crosses the x-axis by finding roots or testing points far from vertex.

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

If the vertex is above the x-axis, why isn't the whole parabola positive?

The vertex is just the highest point of a downward parabola. As you move away from the vertex, the parabola goes down and can cross below the x-axis, creating negative y-values.

How can I tell if a downward parabola crosses the x-axis?

Use the discriminant $b^2 - 4ac$ . If it's positive, the parabola has two real roots and crosses the x-axis. If it's zero, it touches once. If negative, it never crosses.

What's a simple example of this situation?

Consider $y = -x^2 + 1$ . The vertex is at $(0, 1)$ above the x-axis, but when $x = 2$ , we get $y = -4 + 1 = -3$ , which is negative!

Does this apply to upward-opening parabolas too?

No! If an upward-opening parabola has its vertex above the x-axis, then the entire parabola is above the x-axis since that's the lowest point.

How far from the vertex do I need to check?

It depends on the parabola! For $y = -x^2 + c$ , the roots are at $x = ±\sqrt{c}$ . Test points beyond these roots to see negative values.

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