Vertex Intersection with X-Axis: Find X in the Parabolic Function

Q: How can a parabola be positive if it touches the x-axis?

Great question! When a parabola has its vertex on the x-axis, it means the function equals zero at exactly one point. If the parabola opens upward (like a smile), then everywhere else the function is positive!

Q: What does it mean that the vertex is the only intersection point?

This means the parabola touches but doesn't cross the x-axis. It's like a ball bouncing off the ground at exactly one point - it hits the x-axis at the vertex and stays above it everywhere else.

Q: How do I know if the parabola opens up or down?

Look at the coefficient of the squared term! In $ f(x) = a(x - A)^2 $, if a > 0, the parabola opens upward. If a , it opens downward.

Q: Why are both x A correct?

Because the function is positive on both sides of the vertex! The vertex at point A is the only place where $ f(x) = 0 $. Everywhere else (both left and right of A), the function is positive.

Q: What if I can't tell which way the parabola opens?

The problem states that $ f(x) > 0 $ has solutions, which means the parabola must open upward. If it opened downward, the function would be negative everywhere except at the vertex.

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

The graph of the function below intersects the $x$ -axis at point A (the vertex of the parabola).

Find all values of $x$ where $f\left(x\right) > 0$ .

2

Step-by-step solution

To solve this problem, we will look at the behavior of the quadratic function and determine when it is greater than zero:

Step 1: The intersection point A is the vertex, which means $f(x) = a(x - A)^2 + k$ for some constants $a$ and $k=0$ . This implies $f(x)$ changes sign at its vertex.
Step 2: Determine if the parabola opens upwards or downwards. Since the graph of the function intersects the $x$ -axis at the vertex, there are no additional real roots, which indicates either $f(x) \geq 0$ or $f(x) \leq 0$ throughout. As $f(x) > 0$ requires parts of the parabola above the $x$ -axis, the parabola must open upwards.
Step 3: For $f(x) > 0$ , the graph being a parabola indicates positive $x$ intervals are outside of the vertex, i.e., $x < A$ and $x > A$ .
Step 4: The answers fitting this description are (b) $x < A$ and (c) $x > A$ , which combined correspond to option (d) "Answers (b) + (c) are correct".

Therefore, the correct intervals for $f(x) > 0$ are both $x < A$ and $x > A$ , leading to:

Answers (b) + (c) are correct.

3

Final Answer

Answers (b) + (c) are correct.

Key Points to Remember

Essential concepts to master this topic

Vertex Form: When vertex is on x-axis, $f(x) = a(x - A)^2$
Sign Analysis: For $a > 0$ , parabola opens upward making $f(x) > 0$ everywhere except vertex
Check: Test values on both sides: if $A = 3$ , try $x = 2$ and $x = 4$ ✓

Common Mistakes

Avoid these frequent errors

Assuming function is negative everywhere because vertex touches x-axis
Don't think $f(x) \leq 0$ just because the vertex is at (A, 0) = wrong conclusion! The parabola can still open upward, making the function positive everywhere except at the vertex. Always check the direction the parabola opens by examining the coefficient of the squared term.

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

How can a parabola be positive if it touches the x-axis?

+

Great question! When a parabola has its vertex on the x-axis, it means the function equals zero at exactly one point. If the parabola opens upward (like a smile), then everywhere else the function is positive!

What does it mean that the vertex is the only intersection point?

+

This means the parabola touches but doesn't cross the x-axis. It's like a ball bouncing off the ground at exactly one point - it hits the x-axis at the vertex and stays above it everywhere else.

How do I know if the parabola opens up or down?

+

Look at the coefficient of the squared term! In $f(x) = a(x - A)^2$ , if a > 0, the parabola opens upward. If a < 0, it opens downward.

Why are both x < A and x > A correct?

+

Because the function is positive on both sides of the vertex! The vertex at point A is the only place where $f(x) = 0$ . Everywhere else (both left and right of A), the function is positive.

What if I can't tell which way the parabola opens?

+

The problem states that $f(x) > 0$ has solutions, which means the parabola must open upward. If it opened downward, the function would be negative everywhere except at the vertex.

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Vertex Intersection with X-Axis: Find X in the Parabolic Function

Quadratic Functions with Vertex-Only Intersections

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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

How can a parabola be positive if it touches the x-axis?

What does it mean that the vertex is the only intersection point?

How do I know if the parabola opens up or down?

Why are both x < A and x > A correct?

What if I can't tell which way the parabola opens?

🌟 Unlock Your Math Potential

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Parabola Domain Analysis: Finding x-Values Where f(x) > 0