Calculate Area Under (x-3)²: Quadratic Function Analysis

Q: How do I know which side of the vertex is increasing?

For parabolas that open upward like $ y = (x-3)^2 $, the function decreases before the vertex and increases after it. So if vertex is at x = 3, increasing happens when x > 3.

Q: What does 'ascending area' mean?

Ascending area is another way to say increasing interval - the values of x where the function's y-values are getting larger as x increases.

Q: Can I use the derivative to find where it's increasing?

Yes! The derivative of $ y = (x-3)^2 $ is $ y' = 2(x-3) $. When $ y' > 0 $, the function increases. So $ 2(x-3) > 0 $ means $ x > 3 $.

Q: Why isn't the answer x < 3?

Because before the vertex (when x downward! Think of it like a U-shape: the left side goes down, hits the bottom at x = 3, then the right side goes up.

Q: How can I visualize this better?

Try plotting a few points: when x = 1, y = 4; when x = 2, y = 1; when x = 3, y = 0; when x = 4, y = 1; when x = 5, y = 4. See how it decreases until x = 3, then increases?

Quadratic Functions with Increasing Intervals

Find the ascending area of the function

$y=(x-3)^2$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the domain of increase of the function

00:03 We'll use the formula to represent a parabola function

00:08 The formula for intersection point with X-axis (P,K)

00:12 The value P equals (3)

00:15 The value K equals (0)

00:19 Intersection point with X-axis according to the values

00:24 We'll use the shortened multiplication formulas to expand the brackets

00:31 We'll examine the coefficient of X squared, positive - smiling function

00:37 This is also the vertex point of the parabola

00:41 In a minimum parabola, the domain of increase is after the vertex point

00:47 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 ascending area of the function

$y=(x-3)^2$

Step-by-step solution

To determine where the function $y = (x-3)^2$ is increasing, consider the following:

The function is a parabola that opens upwards, centered at the vertex $x = 3$ .
A parabola of the form $y = (x-p)^2$ is increasing on the interval $x > p$ .

This means the function $y = (x-3)^2$ begins to increase after the vertex, which is at $x = 3$ .

Thus, the area of increase (or ascending area) for this function is when $x > 3$ .

Therefore, the correct answer is $3 < x$ .

Final Answer

$3 < x$

Key Points to Remember

Essential concepts to master this topic

Vertex Rule: Parabola $y = (x-h)^2$ has vertex at x = h
Technique: For $y = (x-3)^2$ , vertex is at x = 3
Check: Test values: at x = 4, y = 1; at x = 2, y = 1 but decreasing ✓

Common Mistakes

Avoid these frequent errors

Confusing increasing and decreasing sides of parabola
Don't assume the function increases for x < 3 because the parabola opens upward = wrong interval! This ignores that parabolas decrease before the vertex and increase after. Always remember: for upward parabolas, increasing occurs to the RIGHT of the vertex.

Practice Quiz

Test your knowledge with interactive questions

Find the intersection of the function

$$ y=(x+4)^2 $$

With the Y

FAQ

Everything you need to know about this question

How do I know which side of the vertex is increasing?

For parabolas that open upward like $y = (x-3)^2$ , the function decreases before the vertex and increases after it. So if vertex is at x = 3, increasing happens when x > 3.

What does 'ascending area' mean?

Ascending area is another way to say increasing interval - the values of x where the function's y-values are getting larger as x increases.

Can I use the derivative to find where it's increasing?

Yes! The derivative of $y = (x-3)^2$ is $y' = 2(x-3)$ . When $y' > 0$ , the function increases. So $2(x-3) > 0$ means $x > 3$ .

Why isn't the answer x < 3?

Because before the vertex (when x < 3), the parabola is going downward! Think of it like a U-shape: the left side goes down, hits the bottom at x = 3, then the right side goes up.

How can I visualize this better?

Try plotting a few points: when x = 1, y = 4; when x = 2, y = 1; when x = 3, y = 0; when x = 4, y = 1; when x = 5, y = 4. See how it decreases until x = 3, then increases?

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