Find the Domain of (x+8)²-2¼: Complete Function Analysis

Quadratic Functions with Mixed Number Constants

Find the positive and negative domains of the function below:

y=(x+8)2214 y=\left(x+8\right)^2-2\frac{1}{4}

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Step-by-step written solution

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1

Understand the problem

Find the positive and negative domains of the function below:

y=(x+8)2214 y=\left(x+8\right)^2-2\frac{1}{4}

2

Step-by-step solution

The given function is y=(x+8)2214 y = (x+8)^2 - 2\frac{1}{4} . To find where it is positive or negative, we first find the points where y=0 y = 0 .

Set the function equal to zero:

(x+8)294=0(x+8)^2 - \frac{9}{4} = 0

To handle the fraction, rewrite the equation:

(x+8)2=94(x+8)^2 = \frac{9}{4}

Now, take the square root of both sides:

x+8=±32x+8 = \pm \frac{3}{2}

This gives us two solutions for x x :

x=8+32x = -8 + \frac{3}{2} and x=832x = -8 - \frac{3}{2}

Calculating these values, we have:

x=162+32=132=612x = -\frac{16}{2} + \frac{3}{2} = -\frac{13}{2} = -6\frac{1}{2}

x=16232=192=912x = -\frac{16}{2} - \frac{3}{2} = -\frac{19}{2} = -9\frac{1}{2}

Next, test intervals around the roots to determine where the function is positive or negative:

  • For x<912x < -9\frac{1}{2}, an example point is x=10x = -10: (10+8)294=494(-10+8)^2 - \frac{9}{4} = 4 - \frac{9}{4}, which is positive.
  • For 912<x<612-9\frac{1}{2} < x < -6\frac{1}{2}, an example point is x=8x = -8: (8+8)294=094(-8+8)^2 - \frac{9}{4} = 0 - \frac{9}{4}, which is negative.
  • For x>612x > -6\frac{1}{2}, an example point is x=6x = -6: (6+8)294=494(-6+8)^2 - \frac{9}{4} = 4 - \frac{9}{4}, which is positive.

This leads to the conclusion that:

x<0:912<x<612 x < 0 :-9\frac{1}{2} < x < -6\frac{1}{2}

And for x>0 x > 0 , we have x>612 x > -6\frac{1}{2} or x<0:x<912 x < 0 : x < -9\frac{1}{2}

Thus, the correct answer is:

x<0:912<x<612 x < 0 :-9\frac{1}{2} < x < -6\frac{1}{2}

x>612 x > -6\frac{1}{2} or x>0:x<912 x > 0 : x < -9\frac{1}{2}

3

Final Answer

x<0:912<x<612 x < 0 :-9\frac{1}{2} < x < -6\frac{1}{2}

x>612 x > -6\frac{1}{2} or x>0:x<912 x > 0 : x < -9\frac{1}{2}

Key Points to Remember

Essential concepts to master this topic
  • Domain Analysis: Find zeros first by setting function equal to zero
  • Root Finding: (x+8)2=94 (x+8)^2 = \frac{9}{4} gives x=612,912 x = -6\frac{1}{2}, -9\frac{1}{2}
  • Interval Testing: Check sign in each region: test x=10,8,6 x = -10, -8, -6

Common Mistakes

Avoid these frequent errors
  • Confusing positive/negative domains with function signs
    Don't assume 'positive domain' means x > 0 = wrong intervals! Students often mix up where x-values are positive versus where function values are positive. Always interpret the question as asking where y > 0 (positive) and y < 0 (negative).

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

AAABBBCCCX

FAQ

Everything you need to know about this question

What does 'positive and negative domains' actually mean?

+

This means finding where the function output (y-values) is positive or negative, not where x is positive or negative! You're looking for intervals where y>0 y > 0 and where y<0 y < 0 .

Why do I need to convert the mixed number to an improper fraction?

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Converting 214=94 2\frac{1}{4} = \frac{9}{4} makes the algebra much cleaner! Working with improper fractions prevents calculation errors when taking square roots.

How do I know which intervals to test?

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The zeros x=912 x = -9\frac{1}{2} and x=612 x = -6\frac{1}{2} divide the number line into three regions: left of both zeros, between the zeros, and right of both zeros.

Why is the function positive in two separate intervals?

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Because this is a parabola opening upward! The function is positive outside the roots and negative between them. Think of a U-shape that dips below the x-axis only between the two zeros.

Can I use a graphing calculator to check my answer?

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Absolutely! Graph y=(x+8)2214 y = (x+8)^2 - 2\frac{1}{4} and see where the curve is above the x-axis (positive) and below the x-axis (negative).

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