Analyzing y=x²: Finding Where Parabola Values Stay Positive

Q: Why does (x-2)² equal zero only at x = 2?

Because squaring any real number gives a non-negative result! When x = 2, we get (2-2)² = 0² = 0. For any other x-value, like x = 3, we get (3-2)² = 1² = 1 > 0.

Q: What's the difference between (x-2)² and (x+2)²?

The vertex location changes! $ y = (x-2)^2 $ has vertex at x = 2, while $ y = (x+2)^2 $ has vertex at x = -2. The sign inside the parentheses is opposite to the vertex x-coordinate.

Q: How do I know this parabola opens upward?

The coefficient of the squared term is positive 1! Since there's no negative sign in front of (x-2)², the parabola opens upward, making all y-values ≥ 0.

Q: Could y = -(x-2)² also work for this problem?

No! The negative sign would flip the parabola downward. At x = 2, y would still be 0, but everywhere else y would be negative, not positive as required.

Q: Why isn't y = -x-2 the right answer?

That's a linear function, not a parabola! Linear functions create straight lines, while parabolas are U-shaped curves. Also, y = -x-2 gives negative values in many places, violating the "positive elsewhere" condition.

Quadratic Functions with Vertex Form Transformations

Choose the function that represents the parabola $y=x^2$

positive in all areas except $x=2$ .

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the correct function according to the data

00:03 We'll use the positive domain, understanding that the parabola smiles

00:08 Meaning a positive coefficient for X squared

00:13 We'll draw the function according to the intersection points and type of parabola

00:24 Right shift depends on the P term

00:28 We'll substitute in the parabola formula and solve to find the function

00:37 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Choose the function that represents the parabola $y=x^2$

positive in all areas except $x=2$ .

Step-by-step solution

To solve this problem, we will use the vertex form of a quadratic function, which is $y = (x-h)^2$ , where $h$ is the x-coordinate of the vertex. The problem specifies that the parabola should be zero at $x = 2$ and positive elsewhere, indicating that the vertex of the parabola is at $x = 2$ .

We'll take the following steps to find the correct function:

Step 1: Identify the required vertex: The vertex needs to be at $x = 2$ .
Step 2: Apply this x-coordinate to the vertex form: Using $h = 2$ , the function becomes $y = (x-2)^2$ .
Step 3: Verify that the function meets the condition: At $x = 2$ , $y = (2-2)^2 = 0$ , confirming that the function is zero at this point and positive elsewhere since a square of a real number is non-negative.

Therefore, the solution to the problem is $y = (x-2)^2$ .

Final Answer

$y=(x-2)^2$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: y = (x-h)² places vertex at x = h
Technique: For vertex at x = 2, substitute h = 2 to get y = (x-2)²
Check: At x = 2: y = (2-2)² = 0, elsewhere y > 0 ✓

Common Mistakes

Avoid these frequent errors

Confusing vertex location with function value
Don't think y = (x+2)² has vertex at x = 2 = wrong vertex at x = -2! The vertex form y = (x-h)² means the vertex is at x = h, not x = -h. Always remember that (x-2)² gives vertex at x = 2.

Practice Quiz

Test your knowledge with interactive questions

Find the intersection of the function

$$ y=(x+4)^2 $$

With the Y

FAQ

Everything you need to know about this question

Why does (x-2)² equal zero only at x = 2?

Because squaring any real number gives a non-negative result! When x = 2, we get (2-2)² = 0² = 0. For any other x-value, like x = 3, we get (3-2)² = 1² = 1 > 0.

What's the difference between (x-2)² and (x+2)²?

The vertex location changes! $y = (x-2)^2$ has vertex at x = 2, while $y = (x+2)^2$ has vertex at x = -2. The sign inside the parentheses is opposite to the vertex x-coordinate.

How do I know this parabola opens upward?

The coefficient of the squared term is positive 1! Since there's no negative sign in front of (x-2)², the parabola opens upward, making all y-values ≥ 0.

Could y = -(x-2)² also work for this problem?

No! The negative sign would flip the parabola downward. At x = 2, y would still be 0, but everywhere else y would be negative, not positive as required.

Why isn't y = -x-2 the right answer?

That's a linear function, not a parabola! Linear functions create straight lines, while parabolas are U-shaped curves. Also, y = -x-2 gives negative values in many places, violating the "positive elsewhere" condition.

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