Find the Domain of (x+3¼)² + ¾: Complete Function Analysis

Q: What does 'positive domain' and 'negative domain' actually mean?

Positive domain: all x-values where y > 0 (function is above x-axis)Negative domain: all x-values where y (function is below x-axis). It's about the output values, not the input values!

Q: How do I know this parabola opens upward?

The coefficient of the squared term $ (x+3\frac{1}{4})^2 $ is +1 (positive). When this coefficient is positive, the parabola opens upward like a smile ☺️

Q: Why is the minimum value 3/4 instead of 0?

In vertex form $ (x-h)^2 + k $, the k value represents the vertical shift. Since $ k = \frac{3}{4} $, the entire parabola is shifted up by $ \frac{3}{4} $ units from the standard position.

Q: Can a parabola ever have no negative values?

Absolutely! When an upward-opening parabola has its vertex above the x-axis (k > 0), it never crosses or touches the x-axis. This means $ y > 0 $ for all x-values.

Q: How do I find where x 0 for this function?

This is a trick question! The problem isn't asking about positive/negative x-values. Since $ y > 0 $ for all x, there are no x-values where the function is negative, regardless of whether x itself is positive or negative.

Quadratic Functions with Vertex Form Analysis

Find the positive and negative domains of the function below:

$y=\left(x+3\frac{1}{4}\right)^2+\frac{3}{4}$

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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:

$y=\left(x+3\frac{1}{4}\right)^2+\frac{3}{4}$

Step-by-step solution

To solve this problem, we need to determine when the function $y = \left(x + 3\frac{1}{4}\right)^2 + \frac{3}{4}$ is positive or negative across its domain.

The given function is in the vertex form $(x - h)^2 + k$ , where $h = -3\frac{1}{4}$ and $k = \frac{3}{4}$ . The parabola opens upwards because the coefficient of the squared term is positive.

The vertex of the parabola is at $(-3.25, 0.75)$ . This means the minimum value of $y$ is $0.75$ , which means $y$ is always greater than zero; the function does not reach zero or negative values.

Therefore, the function is positive for all $x$ and non-negative globally. There are no values of $x$ for which the function is negative.

Thus, for this function, the positive domain is all $x$ .

Based on this analysis:

$x < 0 :$ none

$x > 0 :$ for all $x$

Final Answer

$x < 0 :$ none

$x > 0 :$ for all $x$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: $(x-h)^2 + k$ gives vertex at (h,k) directly
Sign Analysis: Since $k = \frac{3}{4} > 0$ , minimum value is positive
Verification: Check vertex $(-3\frac{1}{4}, \frac{3}{4})$ confirms y > 0 always ✓

Common Mistakes

Avoid these frequent errors

Confusing positive/negative domains with positive/negative x-values
Don't think 'positive domain' means x > 0 = wrong interpretation! The question asks where the function y is positive or negative, not about x-values. Always analyze the y-values (function outputs) to determine where the function is above or below the x-axis.

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

What does 'positive domain' and 'negative domain' actually mean?

Positive domain: all x-values where y > 0 (function is above x-axis)
Negative domain: all x-values where y < 0 (function is below x-axis). It's about the output values, not the input values!

How do I know this parabola opens upward?

The coefficient of the squared term $(x+3\frac{1}{4})^2$ is +1 (positive). When this coefficient is positive, the parabola opens upward like a smile ☺️

Why is the minimum value 3/4 instead of 0?

In vertex form $(x-h)^2 + k$ , the k value represents the vertical shift. Since $k = \frac{3}{4}$ , the entire parabola is shifted up by $\frac{3}{4}$ units from the standard position.

Can a parabola ever have no negative values?

Absolutely! When an upward-opening parabola has its vertex above the x-axis (k > 0), it never crosses or touches the x-axis. This means $y > 0$ for all x-values.

How do I find where x < 0 and x > 0 for this function?

This is a trick question! The problem isn't asking about positive/negative x-values. Since $y > 0$ for all x, there are no x-values where the function is negative, regardless of whether x itself is positive or negative.

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Find the Domain of (x+3¼)² + ¾: Complete Function Analysis

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

What does 'positive domain' and 'negative domain' actually mean?

How do I know this parabola opens upward?

Why is the minimum value 3/4 instead of 0?

Can a parabola ever have no negative values?

How do I find where x < 0 and x > 0 for this function?

🌟 Unlock Your Math Potential

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