Find the Domain of y=(x+2)²+12: Positive and Negative Regions

Q: What does 'positive domain' actually mean?

The term is asking for x-values where y is positive, not the domain itself. Since $ y = (x+2)^2 + 12 $ always gives positive y-values, the answer is all real x.

Q: How do I find the minimum value of this parabola?

Use the vertex form! In $ y = (x+2)^2 + 12 $, the vertex is at (-2, 12). Since the coefficient of the squared term is positive, this vertex gives the minimum y-value of 12.

Q: Can I solve this by setting y = 0?

You could try, but $ (x+2)^2 + 12 = 0 $ has no real solutions! This confirms the function is never zero or negative, only positive.

Q: What if the constant term was negative instead?

If we had $ y = (x+2)^2 - 5 $, then the minimum would be -5, creating both positive and negative regions. The sign of the constant term at the vertex determines this!

Quadratic Functions with Positive Domain Analysis

Find the positive and negative domains of the function below:

$y=\left(x+2\right)^2+12$

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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:

$y=\left(x+2\right)^2+12$

Step-by-step solution

To find the positive and negative domains of the function $y = (x+2)^2 + 12$ , follow these steps:

Identify the vertex of the function: The vertex form is $y = (x+2)^2 + 12$ , hence the vertex is at $(-2, 12)$ .
Determine the parabola's direction: Given the coefficient of $(x+2)^2$ is positive, the parabola opens upwards.
Consider the vertex's role: At $x = -2$ , the minimum value of $y$ is 12. Since the parabola opens upwards from there, $y \geq 12$ for all $x$ .
Analyze positivity/negativity: Since the minimum $y$ -value is 12, the function is always positive for all real $x$ , and hence it is not negative for any $x$ .

Therefore, the positive domain is all $x$ , and there is no negative domain. The final choice is:

$x < 0 :$ 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: Identify vertex from y = (x - h)² + k
Minimum Value: At x = -2, y = 12 (lowest point)
Verification: Check y-values: (0+2)² + 12 = 16 > 0 ✓

Common Mistakes

Avoid these frequent errors

Confusing domain with range or positive/negative regions
Don't mix up domain (x-values) with where the function is positive/negative = wrong classification! Domain is all possible x-values, while positive/negative regions describe where y-values are above or below zero. Always analyze the y-output values to determine positive and negative regions.

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

What does 'positive domain' actually mean?

The term is asking for x-values where y is positive, not the domain itself. Since $y = (x+2)^2 + 12$ always gives positive y-values, the answer is all real x.

How do I find the minimum value of this parabola?

Use the vertex form! In $y = (x+2)^2 + 12$ , the vertex is at (-2, 12). Since the coefficient of the squared term is positive, this vertex gives the minimum y-value of 12.

Why is there no negative region for this function?

Because the minimum y-value is 12, and the parabola opens upward from there. Since 12 > 0, all y-values are positive. The function never dips below the x-axis.

Can I solve this by setting y = 0?

You could try, but $(x+2)^2 + 12 = 0$ has no real solutions! This confirms the function is never zero or negative, only positive.

What if the constant term was negative instead?

If we had $y = (x+2)^2 - 5$ , then the minimum would be -5, creating both positive and negative regions. The sign of the constant term at the vertex determines this!

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Find the Domain of y=(x+2)²+12: Positive and Negative Regions

Quadratic Functions with Positive Domain 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

What does 'positive domain' actually mean?

How do I find the minimum value of this parabola?

Why is there no negative region for this function?

Can I solve this by setting y = 0?

What if the constant term was negative instead?

🌟 Unlock Your Math Potential

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