Find the Area Under (x+2)²-1: Quadratic Function Analysis

Quadratic Functions with Decreasing Intervals

Find the descending area of the function

y=(x+2)2−1 y=(x+2)^2-1

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1

Understand the problem

Find the descending area of the function

y=(x+2)2−1 y=(x+2)^2-1

2

Step-by-step solution

To solve this problem, we'll determine where the function y=(x+2)2−1 y = (x+2)^2 - 1 is decreasing. This function is a parabola of the form y=(x−h)2+k y = (x-h)^2 + k , with h=−2 h = -2 and k=−1 k = -1 .

The vertex of this parabola is at (−2,−1) (-2, -1) . Since the coefficient of the squared term (x+2)2(x+2)^2 is positive, the parabola opens upwards. This tells us that the function decreases to the left of the vertex and increases to the right.

Therefore, the function is decreasing for x<−2 x < -2 .

In the context of this multiple-choice question, the decreasing interval corresponds to the choice: x<−2 x < -2 .

Thus, the descending region for the function is x<−2 x < -2 .

3

Final Answer

x<−2 x < -2

Key Points to Remember

Essential concepts to master this topic
  • Vertex Form: y=(x−h)2+k y = (x-h)^2 + k has vertex at (h,k)
  • Parabola Direction: Positive coefficient means opens upward, decreases left of vertex
  • Check: Test x = -3: derivative is negative, confirming decreasing behavior ✓

Common Mistakes

Avoid these frequent errors
  • Confusing vertex coordinates from vertex form
    Don't think the vertex of (x+2)2−1 (x+2)^2 - 1 is at (2, -1)! The vertex form (x−h)2+k (x-h)^2 + k has vertex at (h,k), so (x+2)2 (x+2)^2 means h = -2. Always remember that (x+2)=(x−(−2)) (x+2) = (x-(-2)) , so the vertex x-coordinate is -2.

Practice Quiz

Test your knowledge with interactive questions

Find the corresponding algebraic representation of the drawing:

(0,-4)(0,-4)(0,-4)

FAQ

Everything you need to know about this question

How do I find where a parabola is decreasing?

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For an upward-opening parabola, the function decreases to the left of the vertex and increases to the right. Find the vertex first, then the decreasing interval is everything to the left of that x-coordinate.

What if the parabola opened downward instead?

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If the coefficient of the squared term is negative, the parabola opens downward. Then it would increase to the left of the vertex and decrease to the right - opposite of upward parabolas!

How do I read the vertex from (x+2)2−1 (x+2)^2 - 1 ?

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The vertex form is y=(x−h)2+k y = (x-h)^2 + k with vertex at (h, k). Since we have (x+2)2 (x+2)^2 , this means x−(−2) x - (-2) , so h = -2. The vertex is at (-2, -1).

Why does x<−2 x < -2 mean decreasing?

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At the vertex x = -2, the parabola changes from decreasing to increasing. For any x-value less than -2 (moving left on the graph), the function values get smaller as you move right toward the vertex.

Can I use calculus to check this?

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Yes! The derivative is y′=2(x+2) y' = 2(x+2) . When x<−2 x < -2 , we have y′<0 y' < 0 , confirming the function is decreasing in that interval.

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