Calculate the Descending Area of y=(x-5)² in Quadratic Functions

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

For upward-opening parabolas like $ y = (x-5)^2 $, imagine a U-shape. The function decreases as you go down the left side of the U, then increases as you go up the right side.

Q: What if the parabola opens downward?

If you have a negative coefficient like $ y = -(x-5)^2 $, the parabola opens downward (upside-down U). Then it increases to the left of the vertex and decreases to the right.

Q: Why is the vertex at (5, 0) and not (-5, 0)?

In vertex form $ y = (x-h)^2 + k $, the vertex is at (h, k). For $ y = (x-5)^2 $, we have h = 5 and k = 0, so the vertex is (5, 0).

Q: Can I use calculus to find decreasing intervals?

Yes! Take the derivative: $ y' = 2(x-5) $. The function decreases when \( y' , which gives us \( 2(x-5) , so \( x .

Q: What does 'descending area' mean exactly?

'Descending area' means the same as decreasing interval - the x-values where the function's y-values are getting smaller as x increases from left to right.

Decreasing Intervals with Vertex Form Analysis

Find the descending area of the function

$y=(x-5)^2$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the domain of decrease of the function

00:03 Use shortened multiplication formulas for opening parentheses

00:09 Look at X² coefficient, positive - smiling function

00:15 Look at the function coefficients

00:20 Use the formula to find the vertex

00:25 Substitute appropriate values and solve to find the vertex

00:35 This is the X value at the vertex

00:41 In a maximum parabola, the domain of decrease is after the vertex

00:45 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 descending area of the function

$y=(x-5)^2$

Step-by-step solution

To solve this problem, we will identify where the parabolic function $y = (x-5)^2$ is decreasing.

Step 1: Recognize the vertex of the parabola. The equation $y = (x-5)^2$ indicates the vertex is at $(5, 0)$ .
Step 2: Understand the direction of the parabola. This parabola opens upward as it is in the form $y = (x-p)^2$ .
Step 3: Determine the decreasing interval. For an upward-opening parabola, it decreases on the left side of the vertex.

From the vertex form of the parabola, we can conclude that the function decreases when $x < 5$ .

Therefore, the solution to the problem is $x < 5$ .

Final Answer

$x < 5$

Key Points to Remember

Essential concepts to master this topic

Vertex Rule: For y = (x-h)², vertex is at (h, 0)
Technique: Upward parabolas decrease left of vertex: x < 5
Check: Test x = 3: derivative is negative, confirming decreasing ✓

Common Mistakes

Avoid these frequent errors

Confusing decreasing with increasing intervals
Don't say the function decreases for x > 5 = wrong side of vertex! This gives the increasing interval instead of decreasing. Always remember: upward parabolas decrease to the LEFT 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 decreasing?

For upward-opening parabolas like $y = (x-5)^2$ , imagine a U-shape. The function decreases as you go down the left side of the U, then increases as you go up the right side.

What if the parabola opens downward?

If you have a negative coefficient like $y = -(x-5)^2$ , the parabola opens downward (upside-down U). Then it increases to the left of the vertex and decreases to the right.

Why is the vertex at (5, 0) and not (-5, 0)?

In vertex form $y = (x-h)^2 + k$ , the vertex is at (h, k). For $y = (x-5)^2$ , we have h = 5 and k = 0, so the vertex is (5, 0).

Can I use calculus to find decreasing intervals?

Yes! Take the derivative: $y' = 2(x-5)$ . The function decreases when $y' < 0$ , which gives us $2(x-5) < 0$ , so $x < 5$ .

What does 'descending area' mean exactly?

'Descending area' means the same as decreasing interval - the x-values where the function's y-values are getting smaller as x increases from left to right.

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