Find the Positive Domain: Analyzing y=(x-2)² Function

Q: What does 'positive domain' actually mean?

Positive domain means all x-values where the function gives positive y-values. For $ y = (x-2)^2 $, this is everywhere except x = 2 where y = 0.

Q: Why isn't the answer just x > 2?

Because the parabola is positive on both sides of the vertex! When x = 1, $ y = (1-2)^2 = 1 $, which is positive. The function is positive for all x except x = 2.

Q: How do I know this parabola opens upward?

Look at the coefficient of the squared term! Since $ y = (x-2)^2 $ has a positive coefficient (it's +1), the parabola opens upward like a smile.

Q: What if the parabola opened downward instead?

If it opened downward, like $ y = -(x-2)^2 $, then it would be negative everywhere except the vertex, so the positive domain would be empty or just the single point.

Q: Do I need to expand (x-2)² to solve this?

No! The vertex form $ y = (x-2)^2 $ already shows you the vertex is at x = 2. Expanding to $ y = x^2 - 4x + 4 $ makes it harder to see.

Quadratic Functions with Positive Domain Analysis

What is the positive domain of the function below?

$y=(x-2)^2$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the positive domain of the function

00:03 Positive domain means above the X-axis

00:06 For this, we need to find the intersection points with the X-axis

00:09 At intersection points with X-axis, Y=0, we'll substitute and solve

00:13 Take the square root to eliminate the exponent

00:17 Isolate X

00:24 This is the intersection point with the X-axis

00:32 Use the shortened multiplication formulas and expand the brackets

00:36 Notice the coefficient of X squared is positive

00:39 When the coefficient is positive, the function is smiling

00:45 Draw the function based on intersection points and function type

00:55 The function is positive while it's above the X-axis

01:06 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

What is the positive domain of the function below?

$y=(x-2)^2$

Step-by-step solution

In the first step, we place 0 in place of Y:

0 = (x-2)²

We perform a square root:

0=x-2

x=2

And thus we reveal the point

(2, 0)

This is the vertex of the parabola.

Then we decompose the equation into standard form:

y=(x-2)²

y=x²-4x+2

Since the coefficient of x² is positive, we learn that the parabola is a minimum parabola (smiling).

If we plot the parabola, it seems that it is actually positive except for its vertex.

Therefore the domain of positivity is all X, except X≠2.

Final Answer

all x, $x\ne2$

Key Points to Remember

Essential concepts to master this topic

Vertex Rule: Find where parabola touches x-axis by setting $y = 0$
Technique: $(x-2)^2 = 0$ gives vertex at x = 2
Check: Test values: when x = 3, $y = (3-2)^2 = 1 > 0$ ✓

Common Mistakes

Avoid these frequent errors

Thinking the domain excludes where function is negative
Don't confuse positive domain with where y > 0! Positive domain means all x-values where the function is positive, not the domain itself being positive. Always find where $y > 0$ , which is everywhere except the vertex for this upward parabola.

Practice Quiz

Test your knowledge with interactive questions

Find the intersection of the function

$$ y=(x-2)^2 $$

With the X

FAQ

Everything you need to know about this question

What does 'positive domain' actually mean?

Positive domain means all x-values where the function gives positive y-values. For $y = (x-2)^2$ , this is everywhere except x = 2 where y = 0.

Why isn't the answer just x > 2?

Because the parabola is positive on both sides of the vertex! When x = 1, $y = (1-2)^2 = 1$ , which is positive. The function is positive for all x except x = 2.

How do I know this parabola opens upward?

Look at the coefficient of the squared term! Since $y = (x-2)^2$ has a positive coefficient (it's +1), the parabola opens upward like a smile.

What if the parabola opened downward instead?

If it opened downward, like $y = -(x-2)^2$ , then it would be negative everywhere except the vertex, so the positive domain would be empty or just the single point.

Do I need to expand (x-2)² to solve this?

No! The vertex form $y = (x-2)^2$ already shows you the vertex is at x = 2. Expanding to $y = x^2 - 4x + 4$ makes it harder to see.

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