Find the Area Under y=-(x+2)²+4: Quadratic Function Analysis

Q: How do I know if a parabola opens up or down?

Look at the coefficient a in front of the squared term. If a > 0, the parabola opens upward. If a , it opens downward. In our function $ y = -(x+2)^2 + 4 $, a = -1, so it opens down!

Q: What's the difference between vertex form and standard form?

Vertex form is $ y = a(x-h)^2 + k $ where you can immediately see the vertex at (h,k). Standard form is $ y = ax^2 + bx + c $. Vertex form makes finding increasing/decreasing intervals much easier!

Q: Why is the vertex at (-2, 4) and not (2, 4)?

Be careful with signs! In $ y = -(x+2)^2 + 4 $, we have $ (x-(-2))^2 $, so h = -2. The opposite of what's inside the parentheses gives you the x-coordinate of the vertex.

Q: How do I remember when the function increases vs decreases?

Think of it like a mountain! For a downward parabola: you climb up (increasing) until you reach the peak (vertex), then you go down (decreasing). So increase when x vertex.

Quadratic Functions with Decreasing Intervals

Find the descending area of the function

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

❤️ Continue Your Math Journey!

We have hundreds of course questions with personalized recommendations + Account 100% premium

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the descending area of the function

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

Step-by-step solution

To solve this problem, we need to determine where the function $y = -(x+2)^2 + 4$ is decreasing.

First, note that the function is in the vertex form for a parabola, $y = a(x-h)^2 + k$ . We identify $a = -1$ , $h = -2$ , and $k = 4$ .

The vertex of this parabola is at $(h, k) = (-2, 4)$ . This vertex represents the maximum point of the parabola because $a = -1$ is negative, indicating that the parabola opens downwards.

In a downwards-opening parabola, the function is increasing for values of $x$ less than the vertex $h$ , and it is decreasing for values of $x$ greater than the vertex $h$ .

Therefore, the function is decreasing for $x > -2$ .

Thus, the descending area of the function is:
$-2 < x$ .

The correct choice amongst the provided answers is:

$-2 < x$

Final Answer

$-2 < x$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: y = a(x-h)² + k identifies vertex at (h,k)
Direction Rule: When a < 0, parabola opens downward with maximum at vertex
Decreasing Check: For downward parabola, function decreases when x > vertex x-coordinate ✓

Common Mistakes

Avoid these frequent errors

Confusing increasing and decreasing intervals
Don't assume the function decreases where x < -2 just because the parabola opens downward! This ignores that the vertex at x = -2 is the turning point. Always remember: for downward parabolas, the function increases before the vertex and decreases after the vertex.

Practice Quiz

Test your knowledge with interactive questions

Which equation represents the function:

$$ y=x^2 $$

moved 2 spaces to the right

and 5 spaces upwards.

FAQ

Everything you need to know about this question

How do I know if a parabola opens up or down?

Look at the coefficient a in front of the squared term. If a > 0, the parabola opens upward. If a < 0, it opens downward. In our function $y = -(x+2)^2 + 4$ , a = -1, so it opens down!

What's the difference between vertex form and standard form?

Vertex form is $y = a(x-h)^2 + k$ where you can immediately see the vertex at (h,k). Standard form is $y = ax^2 + bx + c$ . Vertex form makes finding increasing/decreasing intervals much easier!

Why is the vertex at (-2, 4) and not (2, 4)?

Be careful with signs! In $y = -(x+2)^2 + 4$ , we have $(x-(-2))^2$ , so h = -2. The opposite of what's inside the parentheses gives you the x-coordinate of the vertex.

How do I remember when the function increases vs decreases?

Think of it like a mountain! For a downward parabola: you climb up (increasing) until you reach the peak (vertex), then you go down (decreasing). So increase when x < vertex, decrease when x > vertex.

Should I include the vertex point in the decreasing interval?

No! At the vertex, the function is neither increasing nor decreasing - it's at a maximum or minimum. That's why we use $x > -2$ (strict inequality) rather than $x ≥ -2$ .

🌟 Unlock Your Math Potential

Get unlimited access to all 18 Parabola Families questions, detailed video solutions, and personalized progress tracking.

📹

Unlimited Video Solutions

Step-by-step explanations for every problem

📊

Progress Analytics

Track your mastery across all topics

🚫

Ad-Free Learning

Focus on math without distractions

No credit card required • Cancel anytime