Find Negative Area of y+4=(x+6)²: Quadratic Function Analysis

Q: What does 'negative area' mean in this context?

The negative area refers to the region where the function is below the x-axis, meaning y

Q: Why do we use strict inequality < instead of ≤?

Because we want strictly negative values (y

Q: How do I solve (x + 6)² < 4?

Take the square root of both sides to get \( |x + 6| , which means -2 . Then subtract 6 from all parts to get -8

Q: Can I graph this to check my answer?

Absolutely! The parabola $ y = (x + 6)^2 - 4 $ opens upward with vertex at (-6, -4). The function is below the x-axis between the two x-intercepts at x = -8 and x = -4.

Q: What if I forgot to move the 4 to the right side?

You'd be solving $ y + 4 = (x + 6)^2 $ for y Always rewrite in standard form y = (x + 6)² - 4 first to make the inequality easier to solve.

Quadratic Functions with Negative Region Analysis

Find the negative area of the function

$y+4=(x+6)^2$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the negative domain of the function

00:03 Use the abbreviated multiplication formulas and expand the brackets

00:09 Arrange the equation so it describes a function

00:15 Note that the coefficient of X squared is positive

00:24 When the coefficient is positive, the function smiles (opens upward)

00:30 Now we want to find the intersection points with the X-axis

00:36 At the intersection points with X-axis, Y=0, substitute and solve

00:41 Extract the root

00:47 When extracting a root, there are always 2 solutions (positive and negative)

00:50 Solve each possibility to find the points, isolate X

01:11 Draw the function according to intersection points and function type

01:29 The function is negative as long as it's below the X-axis

01:42 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 negative area of the function

$y+4=(x+6)^2$

Step-by-step solution

To find the negative area of the function $y + 4 = (x + 6)^2$ , we need to determine where the function is below the x-axis, i.e., where $y < 0$ .

The equation can be rewritten as:

$y = (x + 6)^2 - 4$ .

We need to solve the inequality:

$(x + 6)^2 - 4 < 0$ .

Adding $4$ to both sides gives:

$(x + 6)^2 < 4$ .

Taking the square root gives:

$-2 < x + 6 < 2$ .

Subtracting $6$ from all sides results in:

$-8 < x < -4$ .

Thus, the interval where the function is below the x-axis is $-8 < x < -4$ .

This corresponds to answer choice 3 from the given options.

Final Answer

$-8 < x < -4$

Key Points to Remember

Essential concepts to master this topic

Rule: Negative area occurs where y < 0 below x-axis
Technique: Solve (x + 6)² < 4 to get -2 < x + 6 < 2
Check: Test x = -6 in interval: y = 0 - 4 = -4 < 0 ✓

Common Mistakes

Avoid these frequent errors

Including boundary points in strict inequality
Don't write -8 ≤ x ≤ -4 for y < 0 = wrong interval type! At x = -8 and x = -4, the function equals zero (y = 0), not negative. Always use strict inequality -8 < x < -4 when finding where y < 0.

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

What does 'negative area' mean in this context?

The negative area refers to the region where the function is below the x-axis, meaning y < 0. It's called 'negative' because the y-values are negative in this interval.

Why do we use strict inequality < instead of ≤?

Because we want strictly negative values (y < 0). At the boundary points x = -8 and x = -4, the function equals zero (y = 0), which is neither positive nor negative.

How do I solve (x + 6)² < 4?

Take the square root of both sides to get $|x + 6| < 2$ , which means -2 < x + 6 < 2. Then subtract 6 from all parts to get -8 < x < -4.

Can I graph this to check my answer?

Absolutely! The parabola $y = (x + 6)^2 - 4$ opens upward with vertex at (-6, -4). The function is below the x-axis between the two x-intercepts at x = -8 and x = -4.

What if I forgot to move the 4 to the right side?

You'd be solving $y + 4 = (x + 6)^2$ for y < 0, which is more complex. Always rewrite in standard form y = (x + 6)² - 4 first to make the inequality easier to solve.

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