Find the Domain of y=-(x-1/3)²+4: Complete Function Analysis

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

The domain is all possible x-values (here: all real numbers). The positive/negative intervals tell you where the function outputs positive or negative y-values.

Q: How do I know if the parabola opens up or down?

Look at the coefficient of the squared term! Since we have $ -1 $ (negative), the parabola opens downward, making it positive between the roots.

Q: Why do we solve when y = 0?

Setting $ y = 0 $ finds the x-intercepts - the boundary points where the function changes from positive to negative (or vice versa).

Q: What if I can't factor the quadratic easily?

Use the square root method like in this problem! Since we have $ (x - \frac{1}{3})^2 = 4 $, take the square root of both sides: $ x - \frac{1}{3} = ±2 $.

Q: How do I remember which intervals are positive?

For a downward parabola: imagine a mountain! The function is positive (above x-axis) between the two roots, and negative (below x-axis) outside the roots.

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

Find the positive and negative domains of the function below:

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

2

Step-by-step solution

To find the positive and negative domains of the function $y = -\left(x - \frac{1}{3}\right)^2 + 4$ , we start by determining where the function crosses the x-axis. This happens where $y = 0$ .

Set $y = 0$ to get:

-\left(x - \frac{1}{3}\right)^2 + 4 = 0

Solving for $x$ :

\left( x - \frac{1}{3} \right)^2 = 4

x - \frac{1}{3} = \pm 2

x = \frac{1}{3} \pm 2

This gives:

Positive root (for $+2$ ):

x = \frac{1}{3} + 2 = \frac{7}{3}

Negative root (for $-2$ ):

x = \frac{1}{3} - 2 = -\frac{5}{3}

The x-intercepts are $x = \frac{7}{3}$ and $x = -\frac{5}{3}$ .

Since the quadratic opens downward (as $a = -1$ ), the graph is above the x-axis between these roots and below outside this interval.

Therefore, the function is positive for:

x \in \left(-\frac{5}{3}, \frac{7}{3}\right)

And negative for:

x > \frac{7}{3}

or

x < -\frac{5}{3}

Thus, the positive domain is:

$x > 0 : -\frac{5}{3} < x < \frac{7}{3}$

And the negative domain is:

$x > \frac{7}{3}$ or $x < 0 : x < -\frac{5}{3}$

The correct choice is:

3

Final Answer

$x > \frac{7}{3}$ or $x < 0 : x < -\frac{5}{3}$

$x > 0 : -\frac{5}{3} < x < \frac{7}{3}$

Key Points to Remember

Essential concepts to master this topic

Domain Rule: Set function equal to zero to find x-intercepts
Technique: Solve $-(x-\frac{1}{3})^2 + 4 = 0$ gives roots at $x = -\frac{5}{3}, \frac{7}{3}$
Check: Since parabola opens downward, function is positive between roots ✓

Common Mistakes

Avoid these frequent errors

Confusing where function is positive versus negative
Don't assume the function is positive when x > 0 and negative when x < 0 = ignores the actual roots! This mixes up the sign of x-values with the sign of y-values. Always find the x-intercepts first, then use the parabola's direction to determine intervals.

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's the difference between domain and positive/negative intervals?

+

The domain is all possible x-values (here: all real numbers). The positive/negative intervals tell you where the function outputs positive or negative y-values.

How do I know if the parabola opens up or down?

+

Look at the coefficient of the squared term! Since we have $-1$ (negative), the parabola opens downward, making it positive between the roots.

Why do we solve when y = 0?

+

Setting $y = 0$ finds the x-intercepts - the boundary points where the function changes from positive to negative (or vice versa).

What if I can't factor the quadratic easily?

+

Use the square root method like in this problem! Since we have $(x - \frac{1}{3})^2 = 4$ , take the square root of both sides: $x - \frac{1}{3} = ±2$ .

How do I remember which intervals are positive?

+

For a downward parabola: imagine a mountain! The function is positive (above x-axis) between the two roots, and negative (below x-axis) outside the roots.

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Find the Domain of y=-(x-1/3)²+4: Complete Function Analysis

Quadratic Functions with Positive-Negative Interval 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's the difference between domain and positive/negative intervals?

How do I know if the parabola opens up or down?

Why do we solve when y = 0?

What if I can't factor the quadratic easily?

How do I remember which intervals are positive?

🌟 Unlock Your Math Potential

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