Solve y=-(x-16)²: Finding Values Where Function is Positive

Q: Why can't this function ever be positive?

Because the function has a negative coefficient in front of the squared term! This means the parabola opens downward, and the vertex at (16,0) is the highest point it can reach.

Q: What does the vertex (16,0) tell me?

The vertex shows the maximum value of the function. Since the highest point is y = 0, the function can never be greater than 0 - it can only equal 0 or be negative.

Q: How do I know which direction the parabola opens?

Look at the coefficient of the squared term! If it's negative (like -1 here), the parabola opens downward. If positive, it opens upward.

Q: What if the question asked for f(x) ≥ 0 instead?

Then the answer would be x = 16 only! The function equals zero at x = 16 and is negative everywhere else, so f(x) ≥ 0 is true only at that single point.

Q: Can I test some x-values to verify this?

At x = 15: y = -(15-16)² = -(-1)² = -1 At x = 17: y = -(17-16)² = -(1)² = -1 At x = 16: y = -(16-16)² = 0This confirms the function is never positive!

Quadratic Functions with Maximum Value Analysis

Look at the function below:

$y=-\left(x-16\right)^2$

Then determine for which values of $x$ the following is true:

$f(x) > 0$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the function below:

$y=-\left(x-16\right)^2$

Then determine for which values of $x$ the following is true:

$f(x) > 0$

Step-by-step solution

To solve this problem, let's analyze the function $y = -\left(x-16\right)^2$ .

The function is in vertex form $y = - (x - 16)^2$ , which suggests it is a quadratic function opening downwards because the coefficient of the squared term is negative.
The vertex of the function is at the point $(16, 0)$ , meaning the maximum point of the parabola is at $y = 0$ .
For a parabola that opens downward, the value of $y$ is always less than or equal to the value at the vertex. Here, since the vertex itself is at 0, there are no values of $x$ for which $y = - (x - 16)^2$ is greater than zero.

Therefore, there are no values of $x$ for which $f(x) > 0$ .

In conclusion, the solution to the problem is: True for no values of $x$ .

Final Answer

True for no values of $x$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: y = -(x-16)² opens downward with vertex at (16,0)
Maximum Point: At vertex x=16, y reaches its highest value of 0
Check Sign: Since coefficient is negative, y ≤ 0 for all x values ✓

Common Mistakes

Avoid these frequent errors

Thinking the function can be positive because it has a vertex
Don't assume y > 0 exists just because there's a clear vertex at (16,0) = wrong conclusion! The negative coefficient means the parabola opens downward, so the vertex is the maximum point at y = 0. Always check if the parabola opens up or down first.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

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

FAQ

Everything you need to know about this question

Why can't this function ever be positive?

Because the function has a negative coefficient in front of the squared term! This means the parabola opens downward, and the vertex at (16,0) is the highest point it can reach.

What does the vertex (16,0) tell me?

The vertex shows the maximum value of the function. Since the highest point is y = 0, the function can never be greater than 0 - it can only equal 0 or be negative.

How do I know which direction the parabola opens?

Look at the coefficient of the squared term! If it's negative (like -1 here), the parabola opens downward. If positive, it opens upward.

What if the question asked for f(x) ≥ 0 instead?

Then the answer would be x = 16 only! The function equals zero at x = 16 and is negative everywhere else, so f(x) ≥ 0 is true only at that single point.

Can I test some x-values to verify this?

At x = 15: y = -(15-16)² = -(-1)² = -1 < 0
At x = 17: y = -(17-16)² = -(1)² = -1 < 0
At x = 16: y = -(16-16)² = 0

This confirms the function is never positive!

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Solve y=-(x-16)²: Finding Values Where Function is Positive

Quadratic Functions with Maximum Value Analysis

❤️ Continue Your Math Journey!

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 this function ever be positive?

What does the vertex (16,0) tell me?

How do I know which direction the parabola opens?

What if the question asked for f(x) ≥ 0 instead?

Can I test some x-values to verify this?

🌟 Unlock Your Math Potential

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