Domain Analysis: Finding Valid Inputs for (x+4)² - 10¼

Quadratic Functions with Positive/Negative Domain Analysis

Find the positive and negative domains of the function below:

y=(x+4)21014 y=\left(x+4\right)^2-10\frac{1}{4}

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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=(x+4)21014 y=\left(x+4\right)^2-10\frac{1}{4}

2

Step-by-step solution

To determine the positive and negative domains of the function, follow these steps:

  • Step 1: Solve (x+4)2=414(x+4)^2 = \frac{41}{4}.
  • Step 2: This implies x+4=±412x+4 = \pm \frac{\sqrt{41}}{2}.
  • Step 3: Solving these gives the roots: x=8+412x = \frac{-8+\sqrt{41}}{2} and x=8412x = \frac{-8-\sqrt{41}}{2}.
  • Step 4: Divide the real number line using these roots into intervals:
    • Interval 1: x<8412x < \frac{-8-\sqrt{41}}{2}
    • Interval 2: 8412<x<8+412\frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2}
    • Interval 3: x>8+412x > \frac{-8+\sqrt{41}}{2}
  • Step 5: Test each interval to see where the function is greater or less than zero, using the sign of (x+4)2414(x+4)^2 - \frac{41}{4}.

Testing reveals that:

  • For Interval 1, the function is positive.
  • For Interval 2, the function is negative.
  • For Interval 3, the function is positive.

Thus, the negative domain is 8412<x<8+412 \frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2} and the positive domains are x>8+412 x > \frac{-8+\sqrt{41}}{2} or x<8412 x < \frac{-8-\sqrt{41}}{2} .

Therefore, the correct answer is:

x<0:8412<x<8+412 x < 0 :\frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2}

x>8+412 x > \frac{-8+\sqrt{41}}{2} or x>0:x<8412 x > 0 : x < \frac{-8-\sqrt{41}}{2}

3

Final Answer

x<0:8412<x<8+412 x < 0 :\frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2}

x>8+412 x > \frac{-8+\sqrt{41}}{2} or x>0:x<8412 x > 0 : x < \frac{-8-\sqrt{41}}{2}

Key Points to Remember

Essential concepts to master this topic
  • Zeros: Set function equal to zero and solve for x-values
  • Technique: Use (x+4)2=1014 (x+4)^2 = 10\frac{1}{4} to find roots x=8±412 x = \frac{-8±\sqrt{41}}{2}
  • Check: Test intervals using sign of (x+4)21014 (x+4)^2 - 10\frac{1}{4}

Common Mistakes

Avoid these frequent errors
  • Testing sign by plugging in x = 0
    Don't just test x = 0 to determine all intervals = completely wrong domains! This only tells you about one point, not entire regions. Always test one value from each interval created by the zeros to determine where the function is positive or negative.

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
\( f\left(x\right) > 0 \).

AAAX

FAQ

Everything you need to know about this question

Why do I need to find where the function equals zero first?

+

The zeros are where the function changes from positive to negative (or vice versa). These critical points divide the number line into intervals where the function keeps the same sign.

How do I convert the mixed number 10¼ to work with it?

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Convert 1014 10\frac{1}{4} to an improper fraction: 414 \frac{41}{4} . This makes solving (x+4)2=414 (x+4)^2 = \frac{41}{4} much easier!

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

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Positive domain: x-values where y > 0 (function is above x-axis)
Negative domain: x-values where y < 0 (function is below x-axis)

How do I test if an interval is positive or negative?

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Pick any test value from each interval and substitute into (x+4)2414 (x+4)^2 - \frac{41}{4} . If result is positive, that interval is in positive domain. If negative, it's in negative domain.

Why are there two separate positive domains?

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Since this is a parabola opening upward, it's positive on both ends and negative in the middle. The vertex form (x+4)21014 (x+4)^2 - 10\frac{1}{4} shows the parabola dips below the x-axis between the two zeros.

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