Parabola Property: Analyzing Vertex Position on X-axis and Negative Values

Q: Why can't we determine if the function is always negative?

Because we don't know the sign of coefficient a in $ y = a(x-h)^2 $. If a > 0, the parabola opens upward (positive values), but if a

Q: What does it mean for the vertex to be on the x-axis?

It means the y-coordinate of the vertex equals zero. So the vertex is at point (h, 0), making the equation $ y = a(x-h)^2 $.

Q: When would the statement be true?

The statement would be true only if a , making the parabola open downward. Then all y-values would be negative except at the vertex where y = 0.

Q: Can a parabola with vertex on x-axis have positive values?

Yes! If a > 0, the parabola opens upward, so all points except the vertex have positive y-values. This contradicts the given statement.

Q: What if a = 0 in the equation?

If a = 0, then $ y = 0(x-h)^2 = 0 $, which is just a horizontal line, not a parabola. We need a ≠ 0 for a true quadratic function.

Parabola Analysis with Vertex Position

If the vertex of the parabola is on the x-axis, then the function is always negative except for the vertex point.

❤️ Continue Your Math Journey!

We have hundreds of course questions with personalized recommendations + Account 100% premium

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

If the vertex of the parabola is on the x-axis, then the function is always negative except for the vertex point.

Step-by-step solution

To solve this problem, let's consider the characteristics of a quadratic function in vertex form:

In vertex form, a quadratic function is written as $y = a(x-h)^2 + k$ . Given that the vertex is on the x-axis, the vertex point, $(h, k)$ , has $k = 0$ . Therefore, the equation becomes $y = a(x-h)^2$ .

We need to determine if the function is always negative except for the vertex point. This boils down to the sign of the coefficient $a$ :

If $a < 0$ , the parabola opens downward, meaning the function is zero at the vertex and negative elsewhere, satisfying the statement.
If $a > 0$ , the parabola opens upward, meaning the function is zero at the vertex and positive elsewhere, contradicting the statement.
If $a = 0$ , the function is linear and doesn't represent a parabola.

Since no specific information regarding the sign of $a$ is given, we cannot conclusively state that the function is always negative except at the vertex.

Therefore, the solution is: Cannot be determined.

Final Answer

Cannot be determined

Key Points to Remember

Essential concepts to master this topic

Rule: Vertex on x-axis means k = 0 in vertex form
Technique: Check sign of coefficient a: negative opens down, positive opens up
Check: Verify both upward and downward cases are possible without more information ✓

Common Mistakes

Avoid these frequent errors

Assuming all parabolas with vertex on x-axis are negative
Don't assume the function is always negative just because the vertex is on the x-axis = wrong conclusion! The sign depends on coefficient a. Always consider both a > 0 (opens up, positive values) and a < 0 (opens down, negative values) cases.

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 can't we determine if the function is always negative?

Because we don't know the sign of coefficient a in $y = a(x-h)^2$ . If a > 0, the parabola opens upward (positive values), but if a < 0, it opens downward (negative values).

What does it mean for the vertex to be on the x-axis?

It means the y-coordinate of the vertex equals zero. So the vertex is at point (h, 0), making the equation $y = a(x-h)^2$ .

When would the statement be true?

The statement would be true only if a < 0, making the parabola open downward. Then all y-values would be negative except at the vertex where y = 0.

Can a parabola with vertex on x-axis have positive values?

Yes! If a > 0, the parabola opens upward, so all points except the vertex have positive y-values. This contradicts the given statement.

What if a = 0 in the equation?

If a = 0, then $y = 0(x-h)^2 = 0$ , which is just a horizontal line, not a parabola. We need a ≠ 0 for a true quadratic function.

🌟 Unlock Your Math Potential

Get unlimited access to all 18 The Quadratic Function questions, detailed video solutions, and personalized progress tracking.

📹

Unlimited Video Solutions

Step-by-step explanations for every problem

📊

Progress Analytics

Track your mastery across all topics

🚫

Ad-Free Learning

Focus on math without distractions

No credit card required • Cancel anytime