Find the Domain of y=-(x+6½)²-2¼: Complete Function Analysis

Quadratic Functions with Domain Classification

Find the positive and negative domains of the function below:

y=(x+612)2214 y=-\left(x+6\frac{1}{2}\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+612)2214 y=-\left(x+6\frac{1}{2}\right)^2-2\frac{1}{4}

2

Step-by-step solution

To solve this problem, we'll determine the domains where the function is positive or negative:

  • Given the function y=(x+612)2214 y = -\left(x + 6\frac{1}{2}\right)^2 - 2\frac{1}{4} , it's in vertex form, with a=1 a = -1 .
  • The vertex is at (x,y)=(612,214) \left(x, y\right) = \left(-6\frac{1}{2}, -2\frac{1}{4}\right) , and since a<0 a < 0 , the parabola opens downward.
  • Because the parabola opens downward and has no x-intercepts due to k<0 k < 0 , the function y=0 y = 0 is never zero nor positive.
  • This implies there are no values of x x for which y>0 y > 0 . Therefore, the positive domain is nonexistent.
  • Thus, the negative domain encompasses all x x such that y0 y \leq 0 , which is the entire set of real numbers.

Since the function's value is never positive, the correct description of positive and negative domains is as follows:
Positive domain: None
Negative domain: For all x x

Therefore, the solution is:

x<0: x < 0 : for all x x

x>0: x > 0 : none

3

Final Answer

x<0: x < 0 : for all x x

x>0: x > 0 : none

Key Points to Remember

Essential concepts to master this topic
  • Vertex Form Analysis: Identify vertex and opening direction from a-value
  • Domain Classification: Separate where y > 0 (positive) and y < 0 (negative)
  • Verification: Check maximum value at vertex: y = -2¼ confirms no positive domain ✓

Common Mistakes

Avoid these frequent errors
  • Confusing domain with range when classifying positive/negative regions
    Don't find where x-values are positive/negative = wrong interpretation! The question asks for x-values where the function OUTPUT (y) is positive or negative. Always analyze the y-values, not x-values.

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

What's the difference between positive domain and negative domain?

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Positive domain: All x-values where y > 0 (function output is positive)
Negative domain: All x-values where y < 0 (function output is negative)

Think of it as: which x-inputs give positive or negative y-outputs?

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

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Look at the coefficient of the squared term (the 'a' value):

  • a > 0: Parabola opens upward (U-shape)
  • a < 0: Parabola opens downward (∩-shape)

In this problem, a = -1, so it opens downward.

Why does this function have no positive domain?

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The parabola opens downward with vertex at (612,214) (-6\frac{1}{2}, -2\frac{1}{4}) . Since the maximum y-value is 214 -2\frac{1}{4} (negative), the function never reaches positive values!

What if the vertex was above the x-axis?

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If the vertex y-coordinate was positive and the parabola opened downward, there would be a positive domain between the two x-intercepts, and negative domains outside those intercepts.

How do I find the vertex from this form?

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In vertex form y=a(xh)2+k y = a(x - h)^2 + k , the vertex is at (h, k).

From y=(x+612)2214 y = -(x + 6\frac{1}{2})^2 - 2\frac{1}{4} :

  • h = 612 -6\frac{1}{2} (opposite sign)
  • k = 214 -2\frac{1}{4}

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