Find the Descending Area: Solving y=(x+1)²+1 Region

Quadratic Functions with Decreasing Intervals

Find the descending area of the function

y=(x+1)2+1 y=(x+1)^2+1

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1

Understand the problem

Find the descending area of the function

y=(x+1)2+1 y=(x+1)^2+1

2

Step-by-step solution

The function given is y=(x+1)2+1 y = (x+1)^2 + 1 . This is a quadratic function, specifically a parabola in vertex form. Let us analyze the function to determine where it is decreasing.

Step 1: Identify the vertex.
The function is in the form y=(x+1)2+1 y = (x+1)^2 + 1 , so the vertex is at the point (1,1) (-1, 1) .

Step 2: Determine the direction in which the parabola opens.
The coefficient of the squared term, (x+1)2 (x+1)^2 , is positive (which is 1), indicating that the parabola opens upwards.

Step 3: Identify where the function is decreasing.
Since the parabola opens upwards, it is decreasing to the left of the vertex and increasing to the right of the vertex. Therefore, the function y=(x+1)2+1 y = (x+1)^2 + 1 is decreasing when x x is less than -1.

Thus, the interval where the function is decreasing is x<1 x < -1 .

The correct answer is: x<1 x < -1 .

3

Final Answer

x<1 x < -1

Key Points to Remember

Essential concepts to master this topic
  • Vertex Form: For y=(x+1)2+1 y = (x+1)^2 + 1 , vertex is at (-1, 1)
  • Parabola Direction: Positive coefficient means opens upward, decreases left of vertex
  • Check: Test x = -2: slope is negative, confirming decreasing for x < -1 ✓

Common Mistakes

Avoid these frequent errors
  • Confusing vertex coordinates with decreasing interval
    Don't think the vertex x-coordinate IS the decreasing interval = wrong boundary! The vertex at x = -1 is just the turning point. Always remember: for upward parabolas, decreasing occurs to the LEFT of the vertex.

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 the vertex from y=(x+1)2+1 y = (x+1)^2 + 1 ?

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In vertex form y=(xh)2+k y = (x-h)^2 + k , the vertex is (h, k). Since we have (x+1)2 (x+1)^2 , that means (x(1))2 (x-(-1))^2 , so h = -1 and k = 1. Vertex is (-1, 1).

Why does the parabola decrease on the left side of the vertex?

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Since the coefficient of (x+1)2 (x+1)^2 is positive (it's 1), the parabola opens upward like a U-shape. This means it goes down until it reaches the bottom (vertex), then goes back up.

What's the difference between 'decreasing' and 'increasing' for functions?

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A function is decreasing when y-values get smaller as x-values get larger (going downward). It's increasing when y-values get larger as x-values get larger (going upward).

Can I use calculus to find where it's decreasing?

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Yes! Take the derivative: y=2(x+1) y' = 2(x+1) . The function decreases when y<0 y' < 0 , so 2(x+1)<0 2(x+1) < 0 gives us x < -1.

How can I verify my answer is correct?

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Pick test values! Try x = -2 (should be decreasing) and x = 0 (should be increasing). At x = -2: y = 2, at x = 0: y = 2. Since the function increases from x = -2 to x = 0, it confirms our interval is right!

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