Domain Analysis: Find Valid Inputs for y=-(x+√13)² Function

Quadratic Functions with Domain Restrictions

Find the positive and negative domains of the function below:

y=(x+13)2 y=-\left(x+\sqrt{1}3\right)^2

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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+13)2 y=-\left(x+\sqrt{1}3\right)^2

2

Step-by-step solution

To find the positive and negative domains of the function, let's first analyze the given function y=(x+13)2 y = -\left(x + \sqrt{13}\right)^2 .

  • Step 1: Analyze the function.
    The function is a downward-opening parabola with vertex at (13,0) (-\sqrt{13}, 0) , because the coefficient of the quadratic term is negative.

  • Step 2: Determine the intervals for positive and negative domains.
    A parabola that opens downward from its vertex means the function is negative for all x13 x \neq -\sqrt{13} since there are no values of x that make the function greater than 0 because the vertex is the maximum point.

  • Step 3: Consider the function around the vertex.
    The only point where the function equals zero is at the vertex x=13 x = -\sqrt{13} . Thus, for any other x x , the function is y<0 y < 0 .

Conclusion:
For x<0 x < 0 , the function satisfies the negative characteristic for all domains except x=13 x = -\sqrt{13} where y=0 y = 0 . Thus, the negative domain is x<0:x13 x < 0 : x \neq -\sqrt{13} .

There are no values of x x for which the function becomes positive. Therefore, the positive domain is empty for x>0 x > 0 .

The solution concludes that the positive domain is none, and the negative domain is x<0:x13 x < 0 : x \neq -\sqrt{13} .

3

Final Answer

x<0:x13 x < 0 : x\ne-\sqrt{13}

x>0: x > 0 : none

Key Points to Remember

Essential concepts to master this topic
  • Rule: Negative parabolas are always non-positive, never positive
  • Technique: Find vertex at x=13 x = -\sqrt{13} where y = 0
  • Check: Test any other x-value: all give negative y-values ✓

Common Mistakes

Avoid these frequent errors
  • Thinking negative parabolas have positive regions
    Don't assume this parabola has positive y-values somewhere = wrong domain analysis! Since the coefficient is negative, the parabola opens downward with maximum value of 0 at the vertex. Always remember negative parabolas are never above the x-axis.

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

Why is there no positive domain for this function?

+

Because the function y=(x+13)2 y = -(x + \sqrt{13})^2 has a negative coefficient in front! This creates a downward-opening parabola that never goes above the x-axis, so y is never positive.

What does the vertex tell us about the function?

+

The vertex at (13,0) (-\sqrt{13}, 0) is the highest point on this parabola. Since the maximum y-value is 0, all other points have negative y-values.

Why do we exclude x = -√13 from the negative domain?

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At x=13 x = -\sqrt{13} , the function equals zero, not negative! We're looking for where y < 0, so we must exclude the point where y = 0.

How do I remember which way a parabola opens?

+

Look at the coefficient of the squared term! If it's positive (+), the parabola opens upward. If it's negative (-), it opens downward.

Could this function ever be positive for any x-value?

+

No! Since this is a downward-opening parabola with its maximum at y = 0, the function can never be positive. The best it can do is reach zero at the vertex.

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