Find the Ascending Region of y=-(x-2)²+1: Quadratic Function Analysis

Q: How do I know which direction a parabola opens?

Look at the coefficient of the squared term! In $ y = -(x-2)^2 + 1 $, the coefficient is -1 (negative), so it opens downward. Positive coefficient = upward, negative = downward.

Q: Why does the function increase for x 2?

Since the parabola opens downward, it reaches its maximum at the vertex (2,1). Moving left from the vertex, the function is climbing up toward this peak, so it's increasing. Moving right, it's falling down, so it's decreasing.

Q: What's the difference between ascending and increasing regions?

They mean the same thing! Ascending region and increasing interval both describe where the function's y-values get larger as x increases. Some textbooks use different terms but the concept is identical.

Q: What if the parabola opened upward instead?

If it opened upward, the increasing region would be to the right of the vertex! Upward parabolas have their minimum at the vertex, so they increase as you move away from the vertex to the right.

Step-by-step video solution

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Step-by-step written solution

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1

Understand the problem

Find the ascending area of the function

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

2

Step-by-step solution

To determine the increasing domain of the function $y = -(x-2)^2 + 1$ , we'll analyze the vertex and the general behavior of parabolas.

Step-by-step solution:

Step 1: Identify the structure of the function.

The given function is $y = -(x-2)^2 + 1$ , which is a quadratic function, specifically a parabola. The general form of a parabola is $y = a(x-h)^2 + k$ . Comparing, we see $a = -1$ , $h = 2$ , and $k = 1$ . The vertex form provides key information about the parabola's orientation, position, and vertex.

Step 2: Analyze the orientation of the parabola.

Since $a = -1$ (which is negative), the parabola opens downwards. A downward-opening parabola indicates that it decreases on either side of the vertex and increases moving towards the vertex from either direction on the x-axis.

Step 3: Determine the vertex's location.

The vertex of the parabola is at $(h, k) = (2, 1)$ . This is the maximum point for this downward-opening parabola since it opens downwards.

Step 4: Find the increasing interval based on the vertex.

For downward-opening parabolas, the interval where the function is increasing is to the left of the vertex. Therefore, the function will be increasing for values less than the x-coordinate of the vertex.

Step 5: Conclusion.

The interval in which the function $y = -(x-2)^2 + 1$ is increasing is $x < 2$ .

Thus, the ascending area (or increasing interval) of the function is $x < 2$ .

3

Final Answer

$x < 2$

Key Points to Remember

Essential concepts to master this topic

Orientation: Negative coefficient means parabola opens downward from vertex
Technique: Vertex at (2,1) from $y = -(x-2)^2 + 1$
Check: Test points: at x=1, y=0; at x=3, y=0; maximum at x=2 ✓

Common Mistakes

Avoid these frequent errors

Confusing increasing direction for downward parabolas
Don't assume the function increases to the right of the vertex = wrong interval! For downward parabolas, the function decreases as you move away from the vertex in both directions. Always remember: downward parabolas increase toward the vertex from the left (x < vertex x-coordinate).

Practice Quiz

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Find the corresponding algebraic representation of the drawing:

FAQ

Everything you need to know about this question

How do I know which direction a parabola opens?

+

Look at the coefficient of the squared term! In $y = -(x-2)^2 + 1$ , the coefficient is -1 (negative), so it opens downward. Positive coefficient = upward, negative = downward.

Why does the function increase for x < 2 instead of x > 2?

+

Since the parabola opens downward, it reaches its maximum at the vertex (2,1). Moving left from the vertex, the function is climbing up toward this peak, so it's increasing. Moving right, it's falling down, so it's decreasing.

What's the difference between ascending and increasing regions?

+

They mean the same thing! Ascending region and increasing interval both describe where the function's y-values get larger as x increases. Some textbooks use different terms but the concept is identical.

How can I visualize this without graphing?

+

Think of walking along the parabola from left to right. For $x < 2$ , you're walking uphill toward the peak at (2,1). For $x > 2$ , you're walking downhill away from the peak.

What if the parabola opened upward instead?

+

If it opened upward, the increasing region would be to the right of the vertex! Upward parabolas have their minimum at the vertex, so they increase as you move away from the vertex to the right.

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Find the Ascending Region of y=-(x-2)²+1: Quadratic Function Analysis

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