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

Q: What if the parabola opened downward instead?

If the coefficient were negative, like $ y = -(x+5)^2 + 5 $, then it would be increasing to the left of the vertex and decreasing to the right. It's the opposite!

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

In vertex form $ y = (x-h)^2 + k $, the vertex is at (h, k). Since we have $ (x+5)^2 $, that's $ (x-(-5))^2 $, so h = -5.

Q: How can I double-check my answer?

Pick a test point! Try x = -6 (which is $ y = (-6+5)^2 + 5 = 6 $. Try x = -4 (which is > -5): $ y = (-4+5)^2 + 5 = 6 $. Since 6 > 5, the function increases away from the vertex in both directions, confirming it decreases for x .

Q: What does 'descending region' mean exactly?

Descending region means the interval where the function values are getting smaller as x increases. For this parabola, y-values decrease as you move from left toward the vertex at x = -5.

Quadratic Function Intervals with Vertex Form

Find the descending area of the function

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

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Understand the problem

Find the descending area of the function

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

Step-by-step solution

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

Step 1: Identify the vertex of the parabola. In the equation $y = (x+5)^2 + 5$ , rewrite as $y = (x - (-5))^2 + 5$ . Thus, the vertex is at $(-5, 5)$ .
Step 2: Recognize that this form represents an upward-opening parabola because the leading coefficient of the quadratic term is positive.
Step 3: The nature of a standard parabola with a positive leading coefficient is that it decreases as $x$ moves from left to right until reaching the vertex, and then it increases.
Step 4: Hence, the function decreases for values of $x$ less than the vertex $x$ -coordinate, which is $-5$ .

Therefore, the function is decreasing for $x < -5$ .

Consequently, the solution is $x < -5$ .

Final Answer

$x < -5$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: In $y = (x+5)^2 + 5$ , vertex is at (-5, 5)
Technique: Upward parabola decreases left of vertex, increases right of vertex
Check: Test x = -6: function decreases from (-6, 6) to (-5, 5) ✓

Common Mistakes

Avoid these frequent errors

Confusing increasing and decreasing regions
Don't assume the function decreases for x > -5 = wrong interval! This confuses left and right sides of the vertex. Always remember: upward parabolas decrease to the LEFT of the vertex and increase to the RIGHT.

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 which side of the vertex is decreasing?

For upward-opening parabolas (positive leading 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 $y = -(x+5)^2 + 5$ , then it would be increasing to the left of the vertex and decreasing to the right. It's the opposite!

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

In vertex form $y = (x-h)^2 + k$ , the vertex is at (h, k). Since we have $(x+5)^2$ , that's $(x-(-5))^2$ , so h = -5.

How can I double-check my answer?

Pick a test point! Try x = -6 (which is < -5): $y = (-6+5)^2 + 5 = 6$ . Try x = -4 (which is > -5): $y = (-4+5)^2 + 5 = 6$ . Since 6 > 5, the function increases away from the vertex in both directions, confirming it decreases for x < -5.

What does 'descending region' mean exactly?

Descending region means the interval where the function values are getting smaller as x increases. For this parabola, y-values decrease as you move from left toward the vertex at x = -5.

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