Find the Domain of Function: (x-4.6)² + 2.1 Analysis

Quadratic Functions with Vertex Form Analysis

Find the positive and negative domains of the function below:

y=(x4.6)2+2.1 y=\left(x-4.6\right)^2+2.1

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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=(x4.6)2+2.1 y=\left(x-4.6\right)^2+2.1

2

Step-by-step solution

Let's determine the positive and negative domains of the quadratic function:

The function given is y=(x4.6)2+2.1 y = (x - 4.6)^2 + 2.1 . This is in the vertex form of a quadratic function y=(xh)2+k y = (x - h)^2 + k .

Key observations:

  • The term (x4.6)2 (x - 4.6)^2 is a square and is thus always non-negative, i.e., it is always 0 \geq 0 .
  • The function y=(x4.6)2+2.1 y = (x - 4.6)^2 + 2.1 describes the value of y y for any x x .
  • The constant +2.1 +2.1 ensures that y y is always positive, specifically y2.1 y \geq 2.1 .

Since the smallest value that (x4.6)2 (x - 4.6)^2 can take is 0, at x=4.6 x = 4.6 , the minimum value of y y is 2.1 2.1 . Thus, for any x x , the output is always positive.

Therefore, we have:

x>0: x > 0 : all x x

x<0: x < 0 : none

This means the function never outputs negative values for any x x .

The correct choice from the given options is:

  • Choice 3: x>0: x > 0 : all x x
  • Choice 3: x<0: x < 0 : none
3

Final Answer

x>0: x > 0 : all x x

x<0: x < 0 : none

Key Points to Remember

Essential concepts to master this topic
  • Domain Rule: All real numbers work for quadratic functions
  • Technique: Minimum value is (4.64.6)2+2.1=2.1 (4.6-4.6)^2 + 2.1 = 2.1
  • Check: Since minimum y-value is 2.1 > 0, function is always positive ✓

Common Mistakes

Avoid these frequent errors
  • Confusing domain with range
    Don't look for where the function equals zero or becomes undefined = wrong focus! This leads to missing that ALL x-values work. Always remember domain means which x-values are allowed as inputs.

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/negative domains and regular domain?

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The domain is all x-values that work (all real numbers here). Positive domain means x-values where y > 0, and negative domain means x-values where y < 0.

Why is the function always positive?

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Because (x4.6)2 (x-4.6)^2 is always ≥ 0 (squares can't be negative), and adding 2.1 makes the minimum value equal to 2.1, which is positive!

How do I find the minimum value quickly?

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In vertex form y=(xh)2+k y = (x-h)^2 + k , the minimum occurs at x = h, and the minimum y-value is k. Here: x = 4.6 gives y = 2.1.

Can a quadratic function have no negative domain?

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Yes! If the parabola opens upward and its vertex is above the x-axis, then y is always positive. The negative domain would be none.

What if the constant was negative instead of +2.1?

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If we had y=(x4.6)23 y = (x-4.6)^2 - 3 , then the minimum would be -3, so the function would have both positive and negative y-values depending on x.

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