Calculate the Ascending Area: Finding Region Above y=-(x+1)²+1

Q: How do I know if a parabola opens up or down?

Look at the coefficient of the squared term! If a > 0, the parabola opens upward (smile). If a , it opens downward (frown). In our function $ y = -(x+1)^2 + 1 $, a = -1, so it opens downward.

Q: Why does the function increase on the left side of the vertex?

Think of climbing a hill! For a downward-opening parabola, the vertex is the highest point. As you move left from the peak, you're going uphill (increasing). As you move right, you're going downhill (decreasing).

Q: How do I find the vertex from the equation?

In vertex form $ y = a(x-h)^2 + k $, the vertex is (h, k). Watch the signs! For $ y = -(x+1)^2 + 1 $, we have (x-(-1)), so h = -1 and k = 1, giving vertex (-1, 1).

Q: Can I use the derivative to find increasing intervals?

Yes! The derivative $ y' = -2(x+1) $ tells us the slope. When $ y' > 0 $, the function increases. So $ -2(x+1) > 0 $ means \( x+1 , giving us \( x .

Q: What if I get confused about which side increases?

Test a point! Pick x = -2 (left of vertex) and x = 0 (right of vertex). Calculate the y-values: at x = -2, y = 0; at x = 0, y = 0. Now try x = -1.5 and x = -0.5 to see which direction goes up!

Quadratic Functions with Increasing Intervals

Find the ascending area of the function

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

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Understand the problem

Find the ascending area of the function

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

Step-by-step solution

To solve this problem, we'll employ the following steps:

Step 1: Identify the function's vertex
Step 2: Determine the parabola's orientation
Step 3: Identify the domain where the function is increasing

Let's proceed with the solution:

Step 1: The function $y = -(x+1)^2 + 1$ is in vertex form, where $a = -1$ , $h = -1$ , and $k = 1$ . Hence, the vertex is at $(-1, 1)$ .

Step 2: Since $a = -1$ , the parabola opens downwards. For a downward-opening parabola, the function increases on the left side of the vertex.

Step 3: The function is increasing for $x$ values less than the vertex's x-coordinate. Therefore, the domain where the function is increasing is $x < -1$ .

In conclusion, the interval where the function $y = -(x+1)^2 + 1$ is increasing is $\boldsymbol{ x < -1 }$ .

Final Answer

$x < -1$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: In $y = a(x-h)^2 + k$ , vertex is at point (h,k)
Direction Rule: When a < 0, parabola opens down and increases left of vertex
Check: Test points: at x = -2, slope is positive; at x = 0, slope is negative ✓

Common Mistakes

Avoid these frequent errors

Confusing increasing with decreasing intervals
Don't assume upward-opening parabolas are the only ones that increase = missing half the problem! Downward-opening parabolas increase on their left side before reaching the peak. Always check the coefficient of the squared term and the vertex location.

Practice Quiz

Test your knowledge with interactive questions

Choose the equation that represents the function

$$ y=-x^2 $$

moved 3 spaces to the left

and 4 spaces up.

FAQ

Everything you need to know about this question

How do I know if a parabola opens up or down?

Look at the coefficient of the squared term! If a > 0, the parabola opens upward (smile). If a < 0, it opens downward (frown). In our function $y = -(x+1)^2 + 1$ , a = -1, so it opens downward.

Why does the function increase on the left side of the vertex?

Think of climbing a hill! For a downward-opening parabola, the vertex is the highest point. As you move left from the peak, you're going uphill (increasing). As you move right, you're going downhill (decreasing).

How do I find the vertex from the equation?

In vertex form $y = a(x-h)^2 + k$ , the vertex is (h, k). Watch the signs! For $y = -(x+1)^2 + 1$ , we have (x-(-1)), so h = -1 and k = 1, giving vertex (-1, 1).

Can I use the derivative to find increasing intervals?

Yes! The derivative $y' = -2(x+1)$ tells us the slope. When $y' > 0$ , the function increases. So $-2(x+1) > 0$ means $x+1 < 0$ , giving us $x < -1$ .

What if I get confused about which side increases?

Test a point! Pick x = -2 (left of vertex) and x = 0 (right of vertex). Calculate the y-values: at x = -2, y = 0; at x = 0, y = 0. Now try x = -1.5 and x = -0.5 to see which direction goes up!

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