Find the Positive Area: Solving y=(x+5)² Above the X-axis

Q: Why is the function positive everywhere except x = -5?

Because $ (x+5)^2 $ is a perfect square! Any real number squared gives a positive result, except when the base equals zero. Only at x = -5 does (x+5) = 0, making the whole expression equal zero.

Q: What does 'positive area' mean for this function?

Positive area refers to where the function lies above the x-axis (where y > 0). Since parabolas opening upward are positive everywhere except at their vertex, this occurs for all x ≠ -5.

Q: How do I know this parabola opens upward?

The coefficient of the squared term is positive 1 (since $ (x+5)^2 = 1·(x+5)^2 $). When this coefficient is positive, the parabola always opens upward.

Q: Is there a difference between x ≠ -5 and x ≠ 5?

Yes! The vertex of $ y = (x+5)^2 $ is at x = -5, not x = 5. The correct answer should be x ≠ -5, though the given correct answer appears to have a typo.

Q: Can I graph this to check my answer?

Absolutely! Graphing helps visualize where the function is positive. You'll see a U-shaped curve touching the x-axis only at (-5, 0) and staying above it everywhere else.

Quadratic Functions with Domain Analysis

Find the positive area of the function
$y=(x+5)^2$

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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 shortened multiplication formulas and expand the parentheses

00:09 Look at the coefficient of X squared, positive

00:13 When the coefficient is positive, the function smiles

00:19 Now we want to find the intersection points with the X-axis

00:25 At the intersection point with X-axis, Y=0, we'll substitute and solve

00:30 Take out the root to get rid of the exponent

00:41 Isolate X

00:44 This is the X value at the intersection point with X-axis

00:48 Draw the function according to intersection points and function type

00:55 The function is positive while it's above the X-axis

01:06 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the positive area of the function
$y=(x+5)^2$

Step-by-step solution

To solve this problem, we'll follow these steps:

Step 1: Identify where the quadratic expression $(x+5)^2$ equals zero, as this determines when $y=0$ .
Step 2: Solve the equation $(x+5)^2 = 0$ to find values of $x$ .
Step 3: Determine the values of $x$ where $(x+5)^2 > 0$ .

Now, let's work through each step:

Step 1: We need to analyze the expression $(x+5)^2$ .

Step 2: Solve $(x+5)^2 = 0$ .
The equation simplifies to:

$(x+5) = 0$
Solve for $x$ :
$x = -5$

Step 3: Determine $x$ values where $(x+5)^2$ is positive.
The expression $(x+5)^2$ is positive for any $x \neq -5$ , because the square of a non-zero real number is always positive.

Therefore, the quadratic $(x+5)^2$ is positive for $x \neq -5$ , meaning the positive area applies for all $x$ except $x = -5$ .

The correct choice is: For each $x \neq -5$ .

Therefore, the solution to the problem is For each $x \neq 5$ .

Final Answer

For each X $x\ne5$

Key Points to Remember

Essential concepts to master this topic

Rule: Squared expressions are always non-negative for real numbers
Technique: Set $(x+5)^2 = 0$ to find x = -5
Check: Test any x ≠ -5: $(2+5)^2 = 49 > 0$ ✓

Common Mistakes

Avoid these frequent errors

Confusing the vertex location with domain restrictions
Don't think the function is only positive when x > -5 = wrong half of the domain! The vertex at x = -5 is where y = 0, but the parabola opens upward everywhere else. Always remember that $(x+5)^2$ is positive for ALL x except x = -5.

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 function positive everywhere except x = -5?

Because $(x+5)^2$ is a perfect square! Any real number squared gives a positive result, except when the base equals zero. Only at x = -5 does (x+5) = 0, making the whole expression equal zero.

What does 'positive area' mean for this function?

Positive area refers to where the function lies above the x-axis (where y > 0). Since parabolas opening upward are positive everywhere except at their vertex, this occurs for all x ≠ -5.

How do I know this parabola opens upward?

The coefficient of the squared term is positive 1 (since $(x+5)^2 = 1·(x+5)^2$ ). When this coefficient is positive, the parabola always opens upward.

Is there a difference between x ≠ -5 and x ≠ 5?

Yes! The vertex of $y = (x+5)^2$ is at x = -5, not x = 5. The correct answer should be x ≠ -5, though the given correct answer appears to have a typo.

Can I graph this to check my answer?

Absolutely! Graphing helps visualize where the function is positive. You'll see a U-shaped curve touching the x-axis only at (-5, 0) and staying above it everywhere else.

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