Find the Domain of (x-2)²+1: Analyzing Positive and Negative Regions

Q: Why does this function never have negative values?

Because squares are always non-negative! The term $ (x-2)^2 $ is always ≥ 0, so adding 1 makes the minimum value equal to 1. Since 1 > 0, the function is always positive.

Q: What's the difference between domain and positive/negative regions?

The domain is all possible x-values (here: all real numbers). The positive region is where y > 0, and negative region is where y

Q: How do I find the vertex quickly?

In vertex form $ y = (x-h)^2 + k $, the vertex is simply (h, k). For $ (x-2)^2 + 1 $, the vertex is (2, 1).

Q: What if the coefficient of the squared term was negative?

If it were $ y = -(x-2)^2 + 1 $, the parabola would open downward and the vertex would be a maximum instead of a minimum. Then you'd need to check if the function goes below zero.

Q: Can I just plug in test values instead?

You could, but finding the vertex is more efficient! Once you know the minimum value is 1, you immediately know the function is always positive without testing multiple points.

Quadratic Functions with Vertex Form Analysis

Find the positive and negative domains of the function below:

$y=\left(x-2\right)^2+1$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the positive and negative domains of the function below:

$y=\left(x-2\right)^2+1$

Step-by-step solution

To solve this problem, we'll analyze the function $y = (x-2)^2 + 1$ step-by-step:

Step 1: Identify the function in its vertex form. The function is already presented as $(x-2)^2 + 1$ , with vertex at $(2, 1)$ .
Step 2: Analyze the minimum point. The vertex represents the minimum value of the quadratic because the coefficient of the squared term, 1, is positive, meaning the parabola opens upwards.
Step 3: Calculate the minimum value. By substituting $x = 2$ into the function, we find $y = (2-2)^2 + 1 = 1$ .
Step 4: Determine the range. Since the minimum value is 1, which is positive, the function never takes negative values. The range of $y$ is $(1, \infty)$ .
Step 5: Establish positive and negative domains: - Negative domain: The function does not have any negative values since it is always at or above 1. - Positive domain: The function is positive for all $x$ , because the minimum value itself (1) is positive.

Therefore:

$x < 0 :$ none

$x > 0 :$ all $x$

Final Answer

$x < 0 :$ none

$x > 0 :$ all $x$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: $(x-h)^2 + k$ gives vertex at (h,k)
Minimum Value: When coefficient is positive, vertex gives minimum $y = 1$
Check Range: Since minimum is 1 > 0, function is always positive ✓

Common Mistakes

Avoid these frequent errors

Confusing where function equals zero with where it's positive/negative
Don't think the function has negative values just because x can be negative! The function $y = (x-2)^2 + 1$ never equals zero or goes below 1. Always check the minimum value first - if it's positive, the entire function is positive.

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 does this function never have negative values?

Because squares are always non-negative! The term $(x-2)^2$ is always ≥ 0, so adding 1 makes the minimum value equal to 1. Since 1 > 0, the function is always positive.

What's the difference between domain and positive/negative regions?

The domain is all possible x-values (here: all real numbers). The positive region is where y > 0, and negative region is where y < 0. This function is positive everywhere!

How do I find the vertex quickly?

In vertex form $y = (x-h)^2 + k$ , the vertex is simply (h, k). For $(x-2)^2 + 1$ , the vertex is (2, 1).

What if the coefficient of the squared term was negative?

If it were $y = -(x-2)^2 + 1$ , the parabola would open downward and the vertex would be a maximum instead of a minimum. Then you'd need to check if the function goes below zero.

Can I just plug in test values instead?

You could, but finding the vertex is more efficient! Once you know the minimum value is 1, you immediately know the function is always positive without testing multiple points.

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Find the Domain of (x-2)²+1: Analyzing Positive and Negative Regions

Quadratic Functions with Vertex Form 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 does this function never have negative values?

What's the difference between domain and positive/negative regions?

How do I find the vertex quickly?

What if the coefficient of the squared term was negative?

Can I just plug in test values instead?

🌟 Unlock Your Math Potential

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