Calculate the Positive Area: Finding Area Under y=(x+6)²

Quadratic Functions with Domain Restrictions

Find the positive area of the function

y=(x+6)2 y=(x+6)^2

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find the positive domain of the function
00:03 Use the shortened multiplication formulas and expand the brackets
00:09 Notice the coefficient of X squared is positive
00:12 When the coefficient is positive, the function smiles
00:18 Now we want to find the intersection points with X-axis
00:22 At intersection points with X-axis, Y=0, substitute and solve
00:31 Take square root to eliminate the power
00:34 Isolate X
00:37 This is the X value at intersection with X-axis
00:42 Let's draw the function according to intersection points and function type:
00:50 The function is positive while it's above the X-axis
00:56 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find the positive area of the function

y=(x+6)2 y=(x+6)^2

2

Step-by-step solution

The given function is y=(x+6)2 y = (x+6)^2 . This function is a parabola open upwards with a vertex at x=−6 x = -6 .

The expression (x+6)2(x+6)^2 signifies the square of a number, which is always non-negative for all real numbers x x . This means (x+6)2≥0(x+6)^2 \geq 0.

To find when the area under the curve is positive, solve for when (x+6)2>0(x+6)^2 > 0. The square of any non-zero number is positive. Therefore, we require:

(x+6)≠0(x+6) \neq 0.

Simplifying this equation, we find:

  • (x+6)=0(x+6) = 0 when x=−6x = -6.
  • Hence, (x+6)2(x+6)^2 is positive wherever x≠−6x \neq -6.

Conclusively, the positive area of this parabola exists at all points except precisely at x=−6x = -6, where the function equals zero.

Looking at the multiple-choice options, the correct answer that aligns with our solution is:

x≠−6 x \neq -6 .

3

Final Answer

x≠−6 x\ne-6

Key Points to Remember

Essential concepts to master this topic
  • Rule: Squared expressions are always non-negative for real numbers
  • Technique: Solve (x+6)2>0 (x+6)^2 > 0 by finding when x+6≠0 x+6 \neq 0
  • Check: At x=−6 x = -6 : (−6+6)2=0 (-6+6)^2 = 0 , everywhere else positive ✓

Common Mistakes

Avoid these frequent errors
  • Confusing when function equals zero vs. when it's positive
    Don't think x≠6 x \neq 6 is correct because you see 6 in the function = wrong vertex location! The vertex is at x=−6 x = -6 where (x+6)2=0 (x+6)^2 = 0 . Always set the expression inside parentheses equal to zero to find the vertex.

Practice Quiz

Test your knowledge with interactive questions

Find the intersection of the function

\( y=(x-2)^2 \)

With the X

FAQ

Everything you need to know about this question

Why is the answer x≠−6 x \neq -6 and not x≠6 x \neq 6 ?

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The function is y=(x+6)2 y = (x+6)^2 , so the vertex occurs when x+6=0 x+6 = 0 , which means x=−6 x = -6 . Don't confuse the sign inside the parentheses!

What does 'positive area' mean for a parabola?

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Positive area means where the function value is greater than zero. Since (x+6)2≥0 (x+6)^2 \geq 0 always, it's positive everywhere except at the single point where it equals zero.

How do I find where a squared expression is positive?

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A squared expression (expression)2 (expression)^2 is positive when the expression inside is not zero. So (x+6)2>0 (x+6)^2 > 0 when x+6≠0 x+6 \neq 0 , meaning x≠−6 x \neq -6 .

Why can't the area be negative for this function?

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Because squares are always non-negative! No matter what value you substitute for x x , you get (x+6)2≥0 (x+6)^2 \geq 0 . The parabola never goes below the x-axis.

Is there only one point where the function equals zero?

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Yes! (x+6)2=0 (x+6)^2 = 0 only when x+6=0 x+6 = 0 , which gives us exactly one solution: x=−6 x = -6 . This is the vertex of the parabola.

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