Calculate the Descending Area of y=(x+4)² : Step-by-Step Guide

Q: What if the parabola opened downward instead?

If the coefficient were negative, like $ y = -(x+4)^2 $, then it would be increasing for $ x and decreasing for \( x > -4 $.

Q: Why is the vertex at (-4, 0) and not (4, 0)?

In vertex form $ y = (x-h)^2 + k $, we have $ y = (x-(-4))^2 + 0 $. The opposite of what's inside gives us h = -4.

Q: How can I verify my answer is correct?

Pick test points! Try x = -5 and x = -3. Calculate: $ f(-5) = 1 $ and $ f(-3) = 1 $. Since -5 f(-3), the function is decreasing as x increases toward -4.

Q: What does 'decreasing area' actually mean?

It means the interval where the function is decreasing - where the y-values get smaller as x-values get larger. It's asking for the x-values, not an actual area measurement!

Quadratic Functions with Decreasing Intervals

Find the descending area of the function

$y=(x+4)^2$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:06 Let's find where the function decreases.

00:11 First, use factoring formulas to expand the brackets.

00:17 Notice the coefficient of X squared is positive. This means it's a smiling function.

00:24 Next, observe the function's coefficients.

00:28 We'll use a formula to find the vertex point.

00:32 Substitute the right values into the formula and solve for the vertex point.

00:41 This gives us the X value at the vertex point.

00:47 In a parabola that opens upward, the function decreases before the vertex point.

00:56 And that's how we solve this problem!

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the descending area of the function

$y=(x+4)^2$

Step-by-step solution

To solve this problem, we will identify and use the vertex of the quadratic function. Knowing that the function $y = (x+4)^2$ is a parabola, we recognize it is in vertex form. The general form of a parabola is $y = (x-h)^2 + k$ , hence the vertex of our function is at $(h, k) = (-4, 0)$ .

Next, it's important to understand the behavior of the function around this vertex. Since the parabola opens upwards (as can be seen from the positive coefficient of the squared term), it will decrease as $x$ moves towards negative infinity, reach its minimum value at the vertex, and then increase as $x$ becomes larger than the vertex.

Therefore, the function is decreasing on the interval to the left of the vertex. This describes when $x < -4$ .

To conclude, the solution to the problem is that the function is decreasing for $x < -4$ .

Final Answer

$x < -4$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: Identify vertex from $y = (x-h)^2 + k$ pattern
Technique: Find vertex at (-4, 0) since $y = (x+4)^2$
Check: Parabola decreases left of vertex, increases right of vertex ✓

Common Mistakes

Avoid these frequent errors

Confusing increasing and decreasing intervals
Don't assume the function decreases where x > -4 = wrong interval! This happens when students confuse left/right movement with up/down behavior. 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

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

With the X

FAQ

Everything you need to know about this question

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

For upward parabolas (positive coefficient), the function always decreases to the left of the vertex and increases to the right. Think of it like a U-shape!

What if the parabola opened downward instead?

If the coefficient were negative, like $y = -(x+4)^2$ , then it would be increasing for $x < -4$ and decreasing for $x > -4$ .

Why is the vertex at (-4, 0) and not (4, 0)?

In vertex form $y = (x-h)^2 + k$ , we have $y = (x-(-4))^2 + 0$ . The opposite of what's inside gives us h = -4.

How can I verify my answer is correct?

Pick test points! Try x = -5 and x = -3. Calculate: $f(-5) = 1$ and $f(-3) = 1$ . Since -5 < -3 but f(-5) > f(-3), the function is decreasing as x increases toward -4.

What does 'decreasing area' actually mean?

It means the interval where the function is decreasing - where the y-values get smaller as x-values get larger. It's asking for the x-values, not an actual area measurement!

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