Analyzing Parabola Properties: Vertex on X-axis with Upward Orientation

Q: Why can't the function be negative if the parabola opens upward?

When a parabola opens upward with $ a > 0 $, the expression $ a(x-h)^2 $ is always non-negative. Since $ (x-h)^2 \geq 0 $ for any real number, and we multiply by positive a, the result cannot be negative!

Q: What happens exactly at the vertex point?

At the vertex $ x = h $, we get $ y = a(h-h)^2 = a \cdot 0 = 0 $. This is the minimum value of the function, where it touches the x-axis exactly once.

Q: How do I know if a parabola opens upward or downward?

Look at the coefficient a in $ y = a(x-h)^2 + k $:If $ a > 0 $, parabola opens upwardIf \( a , parabola opens downward

Q: Can you give me a specific example?

Sure! Consider $ y = 2(x-3)^2 $. The vertex is at (3, 0) on the x-axis. At $ x = 3 $, $ y = 0 $. For any other x-value like $ x = 4 $, we get $ y = 2(1)^2 = 2 > 0 $. Never negative!

Q: What if the statement said the function is always positive except at the vertex?

That would be correct! For an upward-opening parabola with vertex on the x-axis, the function equals zero at the vertex and is positive everywhere else. This matches our mathematical analysis perfectly.

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

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

2

Step-by-step solution

To evaluate the statement about the parabola, we need to understand its properties precisely:

Step 1: Given the parabola's vertex is on the x-axis, we can write its equation in the form $y = a(x-h)^2$ , where $a > 0$ indicates that it opens upwards.

Step 2: The vertex form of a quadratic function has the vertex $(h, 0)$ with the vertex lying directly on the x-axis. Since $a > 0$ , the parabola opens upwards, implying $(h, 0)$ is the minimum point.

Step 3: For points other than the vertex, $(x-h)^2$ is always non-negative. Since $a$ is positive, $y = a(x-h)^2 \geq 0$ . Therefore, the function value at the vertex is zero, and for all other $x$ , the function value is positive.

Step 4: Analyze if the function can be negative: With $a > 0$ , the value $a(x-h)^2$ cannot be negative for any value of $x$ .

Conclusion: The given statement that the function is "always negative except for the vertex point" is incorrect since outside the vertex, the function is always non-negative and can be positive. Hence, the statement is incorrect.

Therefore, the correct answer is Incorrect.

3

Final Answer

Incorrect

Key Points to Remember

Essential concepts to master this topic

Vertex Form: $y = a(x-h)^2$ with vertex at (h, 0)
Sign Analysis: When $a > 0$ , function equals 0 at vertex, positive elsewhere
Verification: Check that $a(x-h)^2 \geq 0$ for all x values ✓

Common Mistakes

Avoid these frequent errors

Assuming upward parabolas with vertex on x-axis are always negative
Don't think that upward-opening parabolas are negative everywhere except the vertex = completely wrong sign analysis! This confuses the minimum value (zero at vertex) with negative values. Always remember that when $a > 0$ and vertex is at (h, 0), the function is zero at the vertex and positive everywhere else.

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 the function be negative if the parabola opens upward?

+

When a parabola opens upward with $a > 0$ , the expression $a(x-h)^2$ is always non-negative. Since $(x-h)^2 \geq 0$ for any real number, and we multiply by positive a, the result cannot be negative!

What happens exactly at the vertex point?

+

At the vertex $x = h$ , we get $y = a(h-h)^2 = a \cdot 0 = 0$ . This is the minimum value of the function, where it touches the x-axis exactly once.

How do I know if a parabola opens upward or downward?

+

Look at the coefficient a in $y = a(x-h)^2 + k$ :

If $a > 0$ , parabola opens upward
If $a < 0$ , parabola opens downward

Can you give me a specific example?

+

Sure! Consider $y = 2(x-3)^2$ . The vertex is at (3, 0) on the x-axis. At $x = 3$ , $y = 0$ . For any other x-value like $x = 4$ , we get $y = 2(1)^2 = 2 > 0$ . Never negative!

What if the statement said the function is always positive except at the vertex?

+

That would be correct! For an upward-opening parabola with vertex on the x-axis, the function equals zero at the vertex and is positive everywhere else. This matches our mathematical analysis perfectly.

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Analyzing Parabola Properties: Vertex on X-axis with Upward Orientation

Parabola Properties with Vertex Analysis

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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 can't the function be negative if the parabola opens upward?

What happens exactly at the vertex point?

How do I know if a parabola opens upward or downward?

Can you give me a specific example?

What if the statement said the function is always positive except at the vertex?

🌟 Unlock Your Math Potential

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