Calculate Negative Area Under y=(x+2)²: Quadratic Function Analysis

Q: Why can't a parabola like $ y = (x+2)^2 $ have negative area?

Because squaring any real number always gives a non-negative result! No matter what value you substitute for x, $ (x+2)^2 $ will be zero or positive, so the parabola never dips below the x-axis.

Q: How do I find the vertex of $ y = (x+2)^2 $ ?

This is in vertex form $ y = (x-h)^2 + k $! Since $ y = (x-(-2))^2 + 0 $, the vertex is at (-2, 0). The parabola's minimum point touches the x-axis here.

Q: What if the question asked about $ y = -(x+2)^2 $ instead?

Great question! With the negative sign, the parabola would open downward with vertex at (-2, 0). Then the entire parabola except the vertex would be below the x-axis, creating negative area.

Q: How can I verify that $ (x+2)^2 \geq 0 $ for all x?

Try any value! For x = 0: $ (0+2)^2 = 4 > 0 $. For x = -5: $ (-5+2)^2 = 9 > 0 $. The only time it equals zero is when x = -2: $ (-2+2)^2 = 0 $.

Quadratic Functions with Non-Negative Values

Find the negative area of the function

$y=(x+2)^2$

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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 shortened multiplication formulas and open parentheses

00:08 Look at the coefficient of X squared, positive

00:12 When the coefficient is positive, the function smiles

00:15 Now we want to find the intersection points with the X-axis

00:20 At the intersection points with X-axis, Y=0, substitute and solve

00:24 Take out the root to get rid of the power

00:27 Isolate X

00:32 This is the X value at the intersection point with X-axis

00:37 Draw the function according to intersection points and function type

00:45 The function is positive while it's above the X-axis

00:51 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

$y=(x+2)^2$

Step-by-step solution

The function $y = (x+2)^2$ describes a parabola that opens upwards and has its vertex at $(-2, 0)$ . Since the equation involves a perfect square, it yields only non-negative values for all $x$ and always lies on or above the x-axis. Therefore, there is no part of this parabola that crosses below the x-axis, resulting in no "negative area" above or below the x-axis.

In conclusion, the correct answer among the choices is: There is no negative area.

Final Answer

There is no

Key Points to Remember

Essential concepts to master this topic

Property: Perfect squares $(x+2)^2$ always yield non-negative values
Analysis: Vertex at $(-2, 0)$ means minimum y-value is 0
Verification: Check y-values: when x = -3, $y = (-3+2)^2 = 1 \geq 0$ ✓

Common Mistakes

Avoid these frequent errors

Assuming negative area exists below x-axis
Don't look for regions where the parabola goes below y = 0! Since $(x+2)^2 \geq 0$ for all real x, the function never produces negative y-values. Always remember that squared expressions are non-negative, so the parabola stays on or above the x-axis.

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 can't a parabola like $y = (x+2)^2$ have negative area?

Because squaring any real number always gives a non-negative result! No matter what value you substitute for x, $(x+2)^2$ will be zero or positive, so the parabola never dips below the x-axis.

What does it mean for a function to have 'negative area'?

Negative area refers to regions where the function's graph lies below the x-axis (where y < 0). Since this parabola has its lowest point at $(-2, 0)$ and opens upward, it never goes below y = 0.

How do I find the vertex of $y = (x+2)^2$ ?

This is in vertex form $y = (x-h)^2 + k$ ! Since $y = (x-(-2))^2 + 0$ , the vertex is at (-2, 0). The parabola's minimum point touches the x-axis here.

What if the question asked about $y = -(x+2)^2$ instead?

Great question! With the negative sign, the parabola would open downward with vertex at (-2, 0). Then the entire parabola except the vertex would be below the x-axis, creating negative area.

How can I verify that $(x+2)^2 \geq 0$ for all x?

Try any value! For x = 0: $(0+2)^2 = 4 > 0$ . For x = -5: $(-5+2)^2 = 9 > 0$ . The only time it equals zero is when x = -2: $(-2+2)^2 = 0$ .

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Calculate Negative Area Under y=(x+2)²: Quadratic Function Analysis

Quadratic Functions with Non-Negative Values

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why can't a parabola like $y = (x+2)^2$ have negative area?

What does it mean for a function to have 'negative area'?

How do I find the vertex of $y = (x+2)^2$ ?

What if the question asked about $y = -(x+2)^2$ instead?

How can I verify that $(x+2)^2 \geq 0$ for all x?

🌟 Unlock Your Math Potential

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

Calculate Negative Area Under y=(x+2)²: Quadratic Function Analysis

Quadratic Functions with Non-Negative Values

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why can't a parabola like y=(x+2)2 y = (x+2)^2 y=(x+2)2 have negative area?

What does it mean for a function to have 'negative area'?

How do I find the vertex of y=(x+2)2 y = (x+2)^2 y=(x+2)2?

What if the question asked about y=−(x+2)2 y = -(x+2)^2 y=−(x+2)2 instead?

How can I verify that (x+2)2≥0 (x+2)^2 \geq 0 (x+2)2≥0 for all x?

🌟 Unlock Your Math Potential

More Questions

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

Why can't a parabola like $y = (x+2)^2$ have negative area?

How do I find the vertex of $y = (x+2)^2$ ?

What if the question asked about $y = -(x+2)^2$ instead?

How can I verify that $(x+2)^2 \geq 0$ for all x?