Increasing and Decreasing Domain of a Parabola: Vertex form of quadratic equation

The function given is $y = -(x+7)^2$, which is a quadratic function in vertex form. The structural form of this function is $y = a(x-h)^2 + k$, where $a = -1$, $h = -7$, and $k = 0$.

The vertex of the parabola is at $(-7, 0)$. Since $a = -1$, the parabola opens downwards. For downward-opening parabolas, the function is increasing to the left of the vertex and decreasing to the right of the vertex.

Therefore, the function is increasing for $x < -7$.

Thus, the solution to the problem is: $x < -7$.

The function given is $y = (x+8)^2 - 1$, which is in vertex form. The vertex of this parabola is at $(-8, -1)$.

Since the coefficient of $(x+8)^2$ is positive ($+1$), the parabola opens upwards. This means that the function is decreasing to the left of the vertex.

In mathematical terms, the function is decreasing when $x$ is less than the vertex $x$-coordinate, $x = -8$.

Therefore, the function is decreasing for the interval $x < -8$.

The function $y = -(x-12)^2 - 4$ is given in vertex form $y = a(x-h)^2 + k$ where $a = -1$, $h = 12$, and $k = -4$. This tells us the vertex of the parabola is at $(12, -4)$. Since $a = -1$ is negative, the parabola opens downward.

In such a parabola, the function is increasing to the left of the vertex and decreasing to the right. The axis of symmetry is $x = 12$. To the left of $x = 12$, the function increases, and to the right of $x = 12$, the function decreases.

Therefore, the function is decreasing when $x > 12$.

Thus, the interval where the function $y = -(x-12)^2 - 4$ is decreasing is for $x > 12$.

The correct answer to this problem is: $x > 12$.

The function $y = (x-8)^2 - 1$ is a quadratic equation in vertex form, indicating a parabola. A parabola in this form $y = a(x-h)^2 + k$ has a vertex at $(h, k)$. For this function, the vertex is located at $(8, -1)$.

Since the coefficient of $(x-8)^2$ is positive (specifically, $a = 1$) the parabola opens upwards. The axis of symmetry is the vertical line $x = 8$, around which the parabola is symmetric. This line divides the parabola into sections where it is decreasing and increasing.

To the left of this vertex (for $x < 8$), the function is decreasing. To the right of this vertex (for $x > 8$), the function is increasing. This is because, as we move away from the vertex on an upward-opening parabola's left side, the y-values decrease.

In conclusion, the interval over which the function is decreasing is $x < 8$.

To determine the intervals where the function $y = (5-x)^2$ is increasing, we follow these steps:

• Step 1: Identify the vertex and axis of symmetry
  The function is given as $y = (5-x)^2$. Rewriting in a standard form gives $y = (x-5)^2$, indicating that this is a standard parabola shifted 5 units to the right.
• Step 2: Determine the direction of the parabola
  The standard form $y = (x-5)^2$ shows a parabola opening downwards. Its vertex is at $x = 5$ and $y = 0$, which serves as the axis of symmetry.
• Step 3: Use calculus for confirmation
  Find the derivative $\frac{dy}{dx} = 2(5-x)(-1) = -2(5-x)$. Simplifying gives $\frac{dy}{dx} = 2(x-5)$. This derivative is zero at the vertex $x = 5$. The function is increasing when $\frac{dy}{dx} > 0$, which occurs when $x > 5$.

From both the vertex and the derivative analysis, the function $y = (5-x)^2$ is increasing when $x > 5$.

Therefore, the interval where the function is increasing is $x > 5$.

To determine the interval where the function $y = (x-4)^2 - 4$ is decreasing, we need to analyze its vertex form and the parabola's properties:

1. The function is in the vertex form $y = (x-4)^2 - 4$, where $(h, k) = (4, -4)$.

2. Since the quadratic function is opening upwards (as $a = 1 > 0$), it means the derivative of the function is negative to the left of the vertex, indicating decreasing behavior for $x < h$.

3. For $y = (x-4)^2 - 4$, the vertex is at $(4, -4)$, so the function is decreasing for all $x < 4$.

Therefore, the interval where the function is decreasing is $x < 4$.

To determine where the function $y = (x+2)^2$ is increasing, follow these steps:

• Step 1: Find the derivative of the function, $y = (x+2)^2$.
  This is $y' = 2(x+2)$.
• Step 2: Identify the critical point by setting the derivative equal to zero: $2(x+2) = 0$.
  The solution to this is $x = -2$, indicating the vertex of the parabola.
• Step 3: Analyze the sign of $y'$ to determine where the function increases.
  If $x > -2$, then $y' = 2(x+2) > 0$, indicating that the function is increasing.
• Thus, the function is increasing for $x > -2$.

Therefore, the interval where the function $y = (x+2)^2$ is increasing is $x > -2$.

To find the intervals where the function $y = (x+10)^2 + 2$ is decreasing, let's proceed as follows:

• Step 1: Recognize that the function is given in vertex form $y = (x+10)^2 + 2$, where the vertex is $(-10, 2)$.
• Step 2: Determine the opening direction of the parabola. Since the coefficient of $(x+10)^2$ is positive (i.e., 1), the parabola opens upwards.
• Step 3: For an upward-opening parabola, the function decreases to the left of the vertex. Thus, as $x$ decreases from $-10$, the function decreases.

Therefore, the function is decreasing in the interval where $x < -10$.

The correct answer is $x < -10$.

To find the intervals where the function $y = (x - 4)^2 - 4$ is increasing, we proceed as follows:

• Step 1: Identify the vertex of the quadratic function. The given function is in vertex form $y = (x - 4)^2 - 4$, where $h = 4$ and $k = -4$. Thus, the vertex is at $(4, -4)$.
• Step 2: Determine the direction of the parabola. The coefficient of the squared term, $a$, is $1$. Since $a > 0$, the parabola opens upwards.
• Step 3: Determine where the function is increasing. For a parabola that opens upwards, the function is increasing to the right of the vertex. Hence, the function is increasing for $x > 4$.

In conclusion, the function $y = (x-4)^2-4$ is increasing for the interval $x > 4$.

Therefore, the solution to the problem is $x > 4$.

To find the intervals where the quadratic function $y = -(x+7)^2 - 5$ is increasing, we will analyze the structure of the function.

The function is in the vertex form $y = a(x-h)^2 + k$. Here, $a = -1$, $h = -7$, and $k = -5$. Therefore, the vertex of this parabola is $(-7, -5)$.

Since $a = -1$, which is less than zero, the parabola opens downward. For parabolas that open downward, the function is increasing on the interval to the left of the vertex and decreasing to the right of the vertex.

Consequently, the function is increasing for $x < -7$.

The correct answer is therefore $x < -7$.

To find the interval where the function $y=-(x+10)^2-4$ is decreasing, we proceed as follows:

• Identify the Vertex: The given function is in the form $y = a(x-h)^2 + k$, where $a = -1$, $h = -10$, and $k = -4$. The vertex of the parabola is $(-10, -4)$.
• Determine the Parabola's Direction: Since $a = -1$, which is less than 0, the parabola opens downwards. This implies the function decreases as we move from the left of the vertex and increases as we move to the right of the vertex.
• Identify the Decreasing Interval: Because the parabola is opening downwards, the function is decreasing on the interval $x > -10$.

Therefore, the function is decreasing on the interval $x > -10$.

The correct multiple-choice answer is $x > -10$.

To identify the intervals of decrease for the function $y = -(x+7)^2 - 5$, we'll analyze its properties:

This function is in the vertex form $y = a(x-h)^2 + k$, where $a = -1$, $h = -7$, and $k = -5$.

• The vertex of this parabola is at $(-7, -5)$.
• The coefficient $a = -1$ tells us that the parabola opens downwards.

For a downward-opening parabola, the function decreases to the right of the vertex. Therefore, the interval where the function is decreasing is when $x > -7$.

Thus, the interval of decrease for the function is $x > -7$.

Therefore, the correct choice for the interval of decrease is $x > -7$.

To identify the intervals where the function $y = -(x+10)^2 - 4$ is increasing, we must first determine the nature and axis of symmetry of this parabola.

Given that the function is in vertex form, $y = a(x-h)^2 + k$, we know the vertex of this parabola is at $(h, k)$. For our function:

• The vertex is $(-10, -4)$ from the formula $y = -(x+10)^2 - 4$.
• Since the coefficient of the quadratic term $a = -1$ is negative, the parabola opens downwards.

For downward-opening parabolas, the function increases to the left of the vertex and decreases to the right of the vertex.

Therefore, the function $y = -(x+10)^2 - 4$ is increasing for values of $x$ less than the x-coordinate of the vertex. Hence, the interval of increase is where $x < -10$.

In conclusion, the interval where the function is increasing is $x < -10$.

To solve this problem and determine the interval where the function $y = (x-5)^2$ is decreasing, follow these steps:

• Step 1: Identify the vertex of the parabola. The function is in vertex form $y = (x-5)^2$, meaning the vertex is at $(5, 0)$.
• Step 2: Understand the behavior of the parabola. This parabola opens upwards because there is no negative sign before the $(x-5)^2$, indicating that it decreases on the interval left of the vertex.
• Step 3: Determine the interval where the function is decreasing. Since the parabola opens upwards, it decreases for $x < 5$.

Therefore, the function $y = (x-5)^2$ is decreasing on the interval $x < 5$.

To solve this problem of finding where the function $y=(x+8)^2-1$ is increasing, we will follow these steps:

• Identify the vertex: The function is in vertex form $y = (x+8)^2 - 1$. Therefore, the vertex is at $(h, k) = (-8, -1)$.
• Determine the direction of opening: Since the coefficient of $(x+8)^2$ is positive (1), the parabola opens upwards.
• Find intervals of increase: For an upward-opening parabola, the function increases to the right of the vertex. Therefore, the function is increasing for $x > -8$.

Thus, the interval where the function is increasing is $x > -8$.

To solve this problem, we'll assess the function $y = (x+2)^2$.

Step 1: Identify the vertex.
The given function is $y = (x+2)^2$, which is in the form $(x-h)^2 + k$. Here, $h = -2$ and $k = 0$, so the vertex is $(-2, 0)$.

Step 2: Determine the orientation of the parabola.
In the expression $y = (x+2)^2$, the coefficient of the square term $a = 1$ is positive, indicating the parabola opens upwards.

Step 3: Identify the decreasing interval.
For a parabola that opens upwards, the function is decreasing to the left of the vertex. The vertex at $x = -2$ marks the transition point from decreasing to increasing.

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

The correct answer is: $x<-2$

To solve this problem, we'll determine where the function $y=-(x-12)^2-4$ is increasing:

First, recognize that this function is a quadratic equation in vertex form:

• The standard form of a quadratic in vertex form is $y=a(x-h)^2+k$.
• In our case, $a = -1$, $h = 12$, and $k = -4$.

Since $a = -1$, which is less than zero, the parabola opens downward. This implies:

• The function is increasing before reaching its vertex.
• The vertex, given by $x = h$, is $x = 12$.

The function is increasing on the interval where $x < 12$ because:

• The function decreases after passing the vertex $x = 12$.
• Therefore, for values of $x$ less than 12, the function increases.

Therefore, the interval where the function is increasing is $x < 12$.

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

• Step 1: Calculate the derivative of the function.
• Step 2: Determine the intervals where the derivative is positive.
• Step 3: Conclude where the function is increasing.

Now, let's work through each step in detail:

Step 1: Calculate the derivative.

Given the function $y = (x+15)^2 + 6$, calculate the derivative:

$y' = \frac{d}{dx}((x+15)^2 + 6) = \frac{d}{dx}(x+15)^2$

Using the power rule: $\frac{d}{dx}(u^2) = 2u \cdot \frac{du}{dx}$, where $u = x+15$, and $\frac{du}{dx} = 1$.

Thus, $y' = 2(x+15) \cdot 1 = 2(x+15)$.

Step 2: Determine where the derivative is positive.

Set the derivative greater than zero to find the increasing interval:

$2(x+15) > 0$.

Divide both sides by 2:

$x+15 > 0$.

Subtract 15 from both sides:

$x > -15$.

Step 3: Conclude where the function is increasing.

The interval where the function is increasing is where $x > -15$.

Therefore, the function $y = (x+15)^2 + 6$ is increasing for $x > -15$.

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

• Step 1: Identify the given function and its form.
• Step 2: Determine the vertex of the quadratic function.
• Step 3: Analyze the function intervals based on the opening direction of the parabola.

Now, let's work through each step:

Step 1: The function is given in the vertex form as $y = -(x-14)^2 - 6$, which identifies $a = -1$, $h = 14$, and $k = -6$.

Step 2: The vertex of this quadratic function is $(14, -6)$.

Step 3: Because $a < 0$ (the coefficient of $(x-h)^2$ is negative), the parabola opens downwards. This means:

• The function is increasing to the left of the vertex, specifically for $x < 14$.
• The function is decreasing to the right of the vertex, specifically for $x > 14$.

Therefore, the function $y = -(x-14)^2 - 6$ is decreasing for $x > 14$.

To solve this problem, we need to determine where the function $y = -(x-14)^2 - 6$ is increasing.

Step 1: Notice that the given function is in the form $y = a(x-h)^2 + k$, which indicates it is a quadratic function.

Step 2: The coefficient $a = -1$, so the parabola opens downwards. This implies the function decreases to the left of the vertex and increases to the right.

Step 3: Identify the vertex of the parabola. The vertex form is $(h, k)$, where $h = 14$ and $k = -6$. Thus, the vertex is at $(14, -6)$.

Step 4: For a downward-opening parabola, the function is increasing on the left side of the vertex. Hence, the function is increasing for $x < 14$.

Therefore, the intervals where the function is increasing is $x < 14$.

The correct answer is: $\textbf{x < 14}$.