Find the Descending Area of y=(x+5)²+2x: Quadratic Function Analysis

Q: Why do I need to expand the function before differentiating?

You don't have to expand first! You can use the chain rule on $ (x+5)^2 $ directly. However, expanding makes the differentiation easier and reduces chances of errors with the chain rule.

Q: What's the difference between decreasing and increasing intervals?

Decreasing: function goes down as x increases (negative slope, y' Increasing: function goes up as x increases (positive slope, y' > 0)

Q: How do I know if I should use < or ≤ in my answer?

Use strict inequality ( for decreasing intervals. At the critical point where y' = 0, the function has zero slope - it's neither increasing nor decreasing at that exact point.

Q: Can I check my answer by graphing?

Absolutely! Graph $ y = (x+5)^2 + 2x $ and visually confirm it's going downward for x

Q: What if I get a different critical point?

Double-check your expansion and derivative. Common errors: forgetting the +2x term when expanding, or arithmetic mistakes. The derivative should be $ y' = 2x + 12 $, giving critical point x = -6.

Quadratic Function Derivatives with Decreasing Intervals

Find the descending area of the function

$y=(x+5)^2+2x$

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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 the shortened multiplication formulas to expand the parentheses

00:11 Collect like terms

00:24 Note the coefficient of X squared, positive - smiling function

00:29 Observe the function's coefficients

00:34 Use the formula to find the vertex point

00:44 Substitute appropriate values and solve to find the vertex point

00:51 This is the X value at the vertex point

00:56 In a minimum parabola, the domain of decrease is before the vertex point

01:01 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

$y=(x+5)^2+2x$

Step-by-step solution

To determine where the function $y = (x+5)^2 + 2x$ is decreasing, we need to follow these steps:

Step 1: Differentiate the Function
The function is $y = (x+5)^2 + 2x$ . First, we expand and simplify it:
$y = (x+5)^2 + 2x = x^2 + 10x + 25 + 2x = x^2 + 12x + 25$ .
Now, compute the derivative:
$y' = \frac{d}{dx} (x^2 + 12x + 25) = 2x + 12$ .
Step 2: Identify Intervals of Decrease
Set the derivative less than zero to find where the function is decreasing:
$2x + 12 < 0$ .
Solve for $x$ :
$2x < -12$
$x < -6$ .

Therefore, the function $y = (x+5)^2 + 2x$ is decreasing for $x < -6$ .

The correct answer is $x < -6$ .

Final Answer

$x < -6$

Key Points to Remember

Essential concepts to master this topic

Rule: Function decreases where derivative is negative (f'(x) < 0)
Technique: Expand $(x+5)^2 + 2x = x^2 + 12x + 25$ , then differentiate
Check: Verify $x = -7: y' = 2(-7) + 12 = -2 < 0$ ✓

Common Mistakes

Avoid these frequent errors

Finding where derivative equals zero instead of less than zero
Don't solve y' = 0 to find decreasing intervals = you'll get critical points only! Setting derivative equal to zero finds where slope is zero (peaks/valleys), not where function decreases. Always solve y' < 0 for decreasing intervals.

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

Why do I need to expand the function before differentiating?

You don't have to expand first! You can use the chain rule on $(x+5)^2$ directly. However, expanding makes the differentiation easier and reduces chances of errors with the chain rule.

What's the difference between decreasing and increasing intervals?

Decreasing: function goes down as x increases (negative slope, y' < 0)
Increasing: function goes up as x increases (positive slope, y' > 0)

How do I know if I should use < or ≤ in my answer?

Use strict inequality (<) for decreasing intervals. At the critical point where y' = 0, the function has zero slope - it's neither increasing nor decreasing at that exact point.

Can I check my answer by graphing?

Absolutely! Graph $y = (x+5)^2 + 2x$ and visually confirm it's going downward for x < -6. The turning point should be at x = -6 where it changes from decreasing to increasing.

What if I get a different critical point?

Double-check your expansion and derivative. Common errors: forgetting the +2x term when expanding, or arithmetic mistakes. The derivative should be $y' = 2x + 12$ , giving critical point x = -6.

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