Analyze x² + 10x + 25: Exploring Positive and Negative Domains

Perfect Square Trinomials with Domain Analysis

Find the positive and negative domains of the function below:

y=x2+10x+25 y=x^2+10x+25

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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=x2+10x+25 y=x^2+10x+25

2

Step-by-step solution

To solve this problem, follow these steps:

  • Step 1: Factor the quadratic function or recognize it as a perfect square trinomial.
  • Step 2: Identify the vertex of the parabola.
  • Step 3: Analyze the intervals between roots and the surrounding areas to determine positivity or negativity.

Now, work through these steps:

Step 1: Recognize the quadratic function y=x2+10x+25 y = x^2 + 10x + 25 as a perfect square trinomial. It can be rewritten as y=(x+5)2 y = (x + 5)^2 .

Step 2: The vertex of the parabola, which also represents its minimum point since the parabola opens upwards, occurs at x=5 x = -5 .

Step 3: Since (x+5)20 (x + 5)^2 \geq 0 for all real x x (because a square is always non-negative), the function is non-negative everywhere.

Therefore, the function is never negative, and since it equals zero at x=5 x = -5 , it is positive for all x5 x \neq -5 .

Therefore, the function's positive domain is x>0:x5 x > 0 : x \neq -5 . For x<0 x < 0 , the function is not negative, hence there is no such domain.

The solution to the problem is:

x>0:x5 x > 0 : x \neq -5

x<0: x < 0 : none

3

Final Answer

x>0:x5 x > 0 :x\ne-5

x<0: x < 0 : none

Key Points to Remember

Essential concepts to master this topic
  • Recognition: x2+10x+25=(x+5)2 x^2 + 10x + 25 = (x + 5)^2 is a perfect square
  • Technique: Since (x+5)20 (x + 5)^2 ≥ 0 always, function is never negative
  • Check: Vertex at x=5 x = -5 gives minimum value of 0 ✓

Common Mistakes

Avoid these frequent errors
  • Confusing positive domain notation with inequality symbols
    Don't write 'x > 0: x ≠ -5' thinking it means x must be greater than 0 = excludes all negative x-values! This misses that the function is positive for ALL x except -5. Always remember that 'positive domain' means where y > 0, not where x > 0.

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 does 'positive domain' actually mean?

+

The positive domain is all x-values where y is positive (y > 0), not where x > 0! Since y=(x+5)2 y = (x + 5)^2 is positive everywhere except at x = -5, the positive domain includes both positive and negative x-values.

Why is x = -5 excluded from the positive domain?

+

At x = -5, we get y=(5+5)2=02=0 y = (-5 + 5)^2 = 0^2 = 0 . Since 0 is neither positive nor negative, x = -5 belongs to neither the positive nor negative domain.

How do I know this function has no negative domain?

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Since (x+5)2 (x + 5)^2 is a perfect square, it's always non-negative (≥ 0). Perfect squares can equal zero but never go below zero, so this function has no negative values.

What's the difference between y = 0 and the negative domain?

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When y = 0, the function touches the x-axis but isn't negative. The negative domain would be where y < 0 (below the x-axis). Since this parabola only touches but never goes below the x-axis, there's no negative domain.

Could the answer be 'all x' for the positive domain?

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No! While the function is non-negative everywhere, it equals exactly zero at x = -5. The positive domain only includes where y > 0, so we must exclude the point where y = 0.

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