Calculate Negative Area: Finding Area Under y=-(x+4)²

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

Negative area refers to regions where the function values are below the x-axis (y

Q: Why is the function negative everywhere except x = -4?

Since $ (x+4)^2 $ is always positive (or zero when x = -4), the negative sign in front makes $ -(x+4)^2 $ always negative or zero. Only at the vertex x = -4 does it equal zero.

Q: How do I find the vertex of this parabola?

For $ y = -(x+4)^2 $, the vertex is at x = -4 because that's where the squared term equals zero. The y-coordinate is y = 0 at this point.

Q: Does this parabola ever go above the x-axis?

No! Since the parabola opens downward with vertex at (−4, 0), the highest point is y = 0. The function is never positive, only negative or zero.

Q: How is this different from y = (x+4)²?

Without the negative sign, $ y = (x+4)^2 $ would open upward and be positive everywhere except at x = -4. The negative sign flips the parabola downward.

Quadratic Functions with Negative Leading Coefficients

Find the negative area of the function

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

❤️ Continue Your Math Journey!

We have hundreds of course questions with personalized recommendations + Account 100% premium

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the domain of negativity in the function

00:03 We'll use the shortened multiplication formulas and open the parentheses

00:09 Negative times positive is always negative

00:19 We'll look at the coefficient of X squared, negative

00:23 When the coefficient is negative, the function is concave down

00:28 Now we want to find the intersection points with the X-axis

00:32 At the intersection point with X-axis, Y=0, we'll substitute and solve

00:39 We'll change from negative to positive

00:43 We'll take the square root to eliminate the exponent

00:48 Isolate X

00:51 This is the X value at the intersection point with X-axis

01:01 We'll draw the function according to the intersection points and function type

01:06 The function is negative while it's below the X-axis

01:11 And this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the negative area of the function

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

Step-by-step solution

To find the negative area for the function $y = -(x+4)^2$ , consider when the function is negative or equals zero:

The parabola $y = -(x+4)^2$ opens downward with the vertex at $x = -4$ , when $x = -4$ , $y = 0$ .
For $x \neq -4$ , the expression $-(x+4)^2$ is negative because $(x + 4)^2$ is always positive or zero, and the negative sign flips it to non-positive, i.e., negative unless zero.
Hence, everywhere except at the vertex $x = -4$ , the function is negative.

The function $y = -(x+4)^2$ is negative for every point except where $x = -4$ .

Therefore, the correct answer from the choice given is: For each $x \neq -4$ .

Final Answer

For each X $x\ne-4$

Key Points to Remember

Essential concepts to master this topic

Sign Rule: Negative coefficient makes parabola open downward everywhere
Technique: At vertex x = -4, y = 0; elsewhere y < 0
Check: Test x = 0: y = -(0+4)² = -16 < 0 ✓

Common Mistakes

Avoid these frequent errors

Thinking negative area means specific x-intervals
Don't assume 'negative area' refers to regions like x < -4 = wrong interpretation! The term means where the function value is negative. Always identify where y < 0, which is everywhere except the vertex for this downward parabola.

Practice Quiz

Test your knowledge with interactive questions

Find the intersection of the function

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

With the Y

FAQ

Everything you need to know about this question

What does 'negative area' actually mean in this context?

Negative area refers to regions where the function values are below the x-axis (y < 0). It's not about specific x-intervals, but about where the graph dips below zero!

Why is the function negative everywhere except x = -4?

Since $(x+4)^2$ is always positive (or zero when x = -4), the negative sign in front makes $-(x+4)^2$ always negative or zero. Only at the vertex x = -4 does it equal zero.

How do I find the vertex of this parabola?

For $y = -(x+4)^2$ , the vertex is at x = -4 because that's where the squared term equals zero. The y-coordinate is y = 0 at this point.

Does this parabola ever go above the x-axis?

No! Since the parabola opens downward with vertex at (−4, 0), the highest point is y = 0. The function is never positive, only negative or zero.

How is this different from y = (x+4)²?

Without the negative sign, $y = (x+4)^2$ would open upward and be positive everywhere except at x = -4. The negative sign flips the parabola downward.

🌟 Unlock Your Math Potential

Get unlimited access to all 18 Parabola Families questions, detailed video solutions, and personalized progress tracking.

📹

Unlimited Video Solutions

Step-by-step explanations for every problem

📊

Progress Analytics

Track your mastery across all topics

🚫

Ad-Free Learning

Focus on math without distractions

No credit card required • Cancel anytime