Parabola Property: Analyzing Positive Values with X-Axis Vertex

Q: What happens if the parabola opens downward instead?

If the parabola opens downward (a negative everywhere except the vertex, which would still be zero. The vertex becomes the highest point instead.

Q: Can I verify this with a specific example?

Yes! Take $ f(x) = (x - 3)^2 $ with vertex at (3, 0). Try x = 2: f(2) = 1 > 0. Try x = 4: f(4) = 1 > 0. Only at x = 3: f(3) = 0.

Q: Does this work for any parabola with vertex on x-axis?

Only for upward-opening parabolas! The key conditions are: vertex on x-axis AND a > 0. Both conditions must be true for the function to be positive everywhere except the vertex.

Q: How do I identify if a parabola opens upward?

Look at the coefficient 'a' in $ f(x) = ax^2 + bx + c $ or $ f(x) = a(x - h)^2 + k $. If a > 0, it opens upward. If a

Parabola Vertex Forms with X-Axis Positioning

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

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Step-by-step written solution

Follow each step carefully to understand the complete solution

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 positive except for the vertex point.

Step-by-step solution

To solve this problem, follow these steps:

Step 1: Recognize that the standard form of a quadratic function is $f(x) = a(x - h)^2 + k$ , which represents a parabola.
Step 2: Since the parabola is upward-opening, the coefficient $a$ is positive, $a > 0$ .
Step 3: Given that the vertex is on the x-axis, the vertex has coordinates $(h, 0)$ , which means $k = 0$ .
Step 4: Substitute $k = 0$ into the standard form to get $f(x) = a(x - h)^2$ .
Step 5: Since $a > 0$ , the expression $a(x - h)^2$ is always non-negative because $(x - h)^2 \geq 0$ for all $x$ .
Step 6: At the vertex $x = h$ , $f(h) = a(h - h)^2 = 0$ ; at any other point, $f(x) > 0$ since $a(x - h)^2 > 0$ .

Therefore, the function value is zero at the vertex and positive everywhere else. This confirms that the statement is correct.

The correct answer to the problem is Correct.

Final Answer

Correct

Key Points to Remember

Essential concepts to master this topic

Vertex Rule: When vertex is on x-axis, k = 0 in $f(x) = a(x - h)^2 + k$
Function Form: Simplifies to $f(x) = a(x - h)^2$ where $a > 0$
Check Values: At vertex f(h) = 0, elsewhere f(x) > 0 since $(x - h)^2 \geq 0$ ✓

Common Mistakes

Avoid these frequent errors

Assuming parabola can have negative values when vertex is on x-axis
Don't think the parabola goes below the x-axis just because it touches it = negative function values! Since the vertex is the minimum point of an upward-opening parabola, all other points must be higher (positive). Always remember that $(x - h)^2 \geq 0$ and when a > 0, the entire function stays non-negative.

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 is the function positive everywhere except the vertex?

When a parabola opens upward and its vertex is on the x-axis, the vertex is the lowest point. Since it touches the x-axis at y = 0, every other point must be above the x-axis, making f(x) > 0 everywhere else.

What happens if the parabola opens downward instead?

If the parabola opens downward (a < 0) with vertex on the x-axis, then the function would be negative everywhere except the vertex, which would still be zero. The vertex becomes the highest point instead.

Can I verify this with a specific example?

Yes! Take $f(x) = (x - 3)^2$ with vertex at (3, 0). Try x = 2: f(2) = 1 > 0. Try x = 4: f(4) = 1 > 0. Only at x = 3: f(3) = 0.

Does this work for any parabola with vertex on x-axis?

Only for upward-opening parabolas! The key conditions are: vertex on x-axis AND a > 0. Both conditions must be true for the function to be positive everywhere except the vertex.

How do I identify if a parabola opens upward?

Look at the coefficient 'a' in $f(x) = ax^2 + bx + c$ or $f(x) = a(x - h)^2 + k$ . If a > 0, it opens upward. If a < 0, it opens downward.

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