Parabola of the Form y=(x-p)²+k: Finding Increasing or Decreasing Domains

To solve this problem, we'll determine where the function $y = (x+2)^2 - 1$ is decreasing. This function is a parabola of the form $y = (x-h)^2 + k$, with $h = -2$ and $k = -1$.

The vertex of this parabola is at $(-2, -1)$. Since the coefficient of the squared term $(x+2)^2$ is positive, the parabola opens upwards. This tells us that the function decreases to the left of the vertex and increases to the right.

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

In the context of this multiple-choice question, the decreasing interval corresponds to the choice: $x < -2$.

Thus, the descending region for the function is $x < -2$.

To solve this problem, we need to determine where the function $y = (x-4)^2 - 2x$ is decreasing. We'll do this by finding where the first derivative is negative. Here's a detailed step-by-step explanation:

• Step 1: Compute the derivative of the function.

We start with the original function:

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

Expanding the square yields:

$y = (x^2 - 8x + 16) - 2x = x^2 - 10x + 16$

Now, let's find the derivative:

$\frac{dy}{dx} = \frac{d}{dx}(x^2 - 10x + 16) = 2x - 10$

• Step 2: Determine the values where the derivative is negative.

We need to solve the inequality:

$2x - 10 < 0$

Solving for $x$:

$2x < 10$

$x < 5$

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

Conclusion: The function $y = (x-4)^2 - 2x$ is decreasing for $x < 5$.

The correct choice from the options given is therefore:

$x < 5$

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

The function given is $y = (x+1)^2 + 1$. This is a quadratic function, specifically a parabola in vertex form. Let us analyze the function to determine where it is decreasing.

Step 1: Identify the vertex.
The function is in the form $y = (x+1)^2 + 1$, so the vertex is at the point $(-1, 1)$.

Step 2: Determine the direction in which the parabola opens.
The coefficient of the squared term, $(x+1)^2$, is positive (which is 1), indicating that the parabola opens upwards.

Step 3: Identify where the function is decreasing.
Since the parabola opens upwards, it is decreasing to the left of the vertex and increasing to the right of the vertex. Therefore, the function $y = (x+1)^2 + 1$ is decreasing when $x$ is less than -1.

Thus, the interval where the function is decreasing is x < -1 .

The correct answer is: x < -1 .

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 $x < -5$.

To solve this problem, we'll follow these steps:

• Step 1: Identify the vertex of the function.
• Step 2: Determine the direction in which the parabola opens.
• Step 3: Use the vertex to determine where the function is increasing.

Now, let's work through each step:
Step 1: The function $y = (x+4)^2 - 5$ is in vertex form, with the vertex at $x = -4$.
Step 2: The expression $(x+4)^2$ has a positive coefficient, indicating the parabola opens upwards.
Step 3: For an upwards opening parabola, the function increases for values of $x$ greater than the vertex. Therefore, the function is increasing for $x > -4$.

Therefore, the solution to the problem is $-4 < x$.

To solve this problem, we'll employ the following steps:

• Step 1: Identify the function's vertex
• Step 2: Determine the parabola's orientation
• Step 3: Identify the domain where the function is increasing

Let's proceed with the solution:

Step 1: The function $y = -(x+1)^2 + 1$ is in vertex form, where $a = -1$, $h = -1$, and $k = 1$. Hence, the vertex is at $(-1, 1)$.

Step 2: Since $a = -1$, the parabola opens downwards. For a downward-opening parabola, the function increases on the left side of the vertex.

Step 3: The function is increasing for $x$ values less than the vertex's x-coordinate. Therefore, the domain where the function is increasing is $x < -1$.

In conclusion, the interval where the function $y = -(x+1)^2 + 1$ is increasing is $\boldsymbol{ x < -1 }$.

The function given is $y = (x-6)^2 + 3$. This is a standard parabola with vertex form $(x - p)^2 + k$.

Step-by-step solution:

• Identify the vertex: The vertex form of a parabola is given as $y = (x-p)^2 + k$, where $(p, k)$ is the vertex. For this function, $p = 6$ and $k = 3$, so the vertex is $(6, 3)$.
• Determine the direction of the parabola: The parabola opens upwards because the coefficient of $(x-6)^2$ is positive (specifically, it equals 1). This means the parabola decreases on the left of the vertex and increases on the right.
• Determine the domain for increasing values: Since the vertex is at $x = 6$ and the parabola opens upwards, the graph of the function is increasing for all $x$ values greater than 6. Thus, for $x > 6$, the function $y = (x-6)^2 + 3$ is increasing.

Therefore, the solution to the problem is $x > 6$.

Thus, the correct answer from the given choices is: $6 < x$.

To solve this problem, we'll identify the vertex of the given parabolic function and determine on which side of the vertex the function decreases.

Step 1: Identify the Vertex
The function $y = -(x+3)^2 - 10$ is in the vertex form $y = a(x-h)^2 + k$. Here, $a = -1$, $h = -3$, and $k = -10$. Thus, the vertex of the parabola is at $(-3, -10)$.

Step 2: Analyze Parabola's Direction
Since $a = -1$ (a negative value), the parabola opens downwards. For downward-opening parabolas, the function decreases to the right of its vertex.

Step 3: Determine the Decreasing Domain
Since the parabola decreases for values of $x$ greater than the x-coordinate of the vertex, it decreases when $x > -3$.

Therefore, the descending area of the function is $-3 < x$.

The correct answer, corresponding to the choices given, is choice 3: $-3 < x$.

To determine the increasing domain of the function $y = -(x-2)^2 + 1$, we'll analyze the vertex and the general behavior of parabolas.

Step-by-step solution:

• Step 1: Identify the structure of the function.

The given function is $y = -(x-2)^2 + 1$, which is a quadratic function, specifically a parabola. The general form of a parabola is $y = a(x-h)^2 + k$. Comparing, we see $a = -1$, $h = 2$, and $k = 1$. The vertex form provides key information about the parabola's orientation, position, and vertex.

• Step 2: Analyze the orientation of the parabola.

Since $a = -1$ (which is negative), the parabola opens downwards. A downward-opening parabola indicates that it decreases on either side of the vertex and increases moving towards the vertex from either direction on the x-axis.

• Step 3: Determine the vertex's location.

The vertex of the parabola is at $(h, k) = (2, 1)$. This is the maximum point for this downward-opening parabola since it opens downwards.

• Step 4: Find the increasing interval based on the vertex.

For downward-opening parabolas, the interval where the function is increasing is to the left of the vertex. Therefore, the function will be increasing for values less than the x-coordinate of the vertex.

• Step 5: Conclusion.

The interval in which the function $y = -(x-2)^2 + 1$ is increasing is $x < 2$.

Thus, the ascending area (or increasing interval) of the function is $x < 2$.

To solve this problem, we first note that the function is expressed in vertex form as $y = (x-6)^2 + 4$. The vertex form of a quadratic function is $y = a(x - p)^2 + k$, where the vertex is located at $(p, k)$. For our function, this vertex is at $(6, 4)$.

Since the coefficient of $(x-6)^2$ is positive, the parabola opens upwards. This means that the lowest point on the parabola is the vertex, making $x = 6$ the axis of symmetry and the point at which the function changes from decreasing to increasing.

Thus, the function is increasing on the interval where $x > 6$. When $x$ is greater than 6, the value of $y$ increases as $x$ increases.

By examining the multiple-choice answers, we find that the choice that matches $6 < x$ is the correct one. Therefore, the ascending area of the function is for $x > 6$.

Therefore, the solution to the problem is $6 < x$.