Vertex Representation: Finding Increasing or Decreasing Domains

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 $x < 8$.

The given function is $y = -(x+\frac{7}{8})^2 - 1\frac{1}{5}$. This function is in the vertex form $y = a(x-h)^2 + k$, where $a = -1$, $h = -\frac{7}{8}$, and $k = -\frac{6}{5}$.

• The vertex of the quadratic function is $(- \frac{7}{8}, -\frac{6}{5})$.
• Since $a = -1$ which is less than 0, the parabola opens downward.
• For a parabola that opens downward, the function is increasing on the interval where $x < h$ and decreasing on the interval where $x > h$.
• Thus, the function increasing on $(- \infty, -\frac{7}{8})$ and decreasing on $(- \frac{7}{8}, \infty)$.

Therefore, the correct intervals are:
$\nearrow: x < -\frac{7}{8}$
$\searrow: x > -\frac{7}{8}$

Therefore, the final intervals of increase and decrease are:

$\searrow: x > -\frac{7}{8}$
$\nearrow: x < -\frac{7}{8}$

The given quadratic function is $y = -\left(x - \frac{1}{3}\right)^2 + 4$. This is in vertex form, where the vertex is $\left(\frac{1}{3}, 4\right)$ and the coefficient $a = -1$ indicates the parabola opens downward.

For a downward-opening parabola, the function decreases immediately after the vertex and increases before it. Thus, we identify:

• The function is increasing for $x < \frac{1}{3}$.
• The function is decreasing for $x > \frac{1}{3}$.

Therefore, the intervals of increase and decrease of the function are:

$\searrow: x > \frac{1}{3}$ (decreasing) $\nearrow: x < \frac{1}{3}$ (increasing)

The correct answer, corresponding to the choices given, is:

$\searrow:x>\frac{1}{3}\\\nearrow:x<\frac{1}{3}$

To determine the intervals of increase and decrease for the function $y = \left(x - 8\frac{1}{6}\right)^2 - 1$, we'll follow these steps:

• Identify the vertex of the parabola.
• Determine the direction the parabola opens (upwards for $a > 0$).
• Determine intervals based on the axis of symmetry at the vertex.

Step 1: Identify the vertex. The given function is in vertex form $y = (x - h)^2 + k$, where $h = 8\frac{1}{6}$ and $k = -1$. Thus, the vertex is $\left(8\frac{1}{6}, -1\right)$.

Step 2: Determine direction. The coefficient $a = 1$ is positive, so the parabola opens upwards. This implies the function decreases before the vertex and increases after the vertex.

Step 3: Determine intervals of increase and decrease. Since the parabola reaches a minimum at the vertex $x = 8\frac{1}{6}$:
- The function is decreasing for $x < 8\frac{1}{6}$.
- The function is increasing for $x > 8\frac{1}{6}$.

Therefore, the intervals of increase and decrease are as follows:
Decreasing interval: $x < 8\frac{1}{6}$.
Increasing interval: $x > 8\frac{1}{6}$.

The correct answer is:
$\searrow:x<8\frac{1}{6}\\\nearrow:x>8\frac{1}{6}$.

To determine the intervals of increase and decrease for the function $y = -\left(x-\frac{4}{9}\right)^2 + 1$, we follow these steps:

• Identify Vertex: The function is in vertex form $y = a(x-h)^2 + k$, where the vertex is at $\left(\frac{4}{9}, 1 \right)$.
• Parabola Direction: Since $a = -1$, which is negative, the parabola opens downwards.
• Interval Analysis:
  • The function is increasing on the interval $x < \frac{4}{9}$, moving left towards the vertex.
  • The function is decreasing on the interval $x > \frac{4}{9}$, moving right away from the vertex.

Therefore, after analyzing the function's behavior, we find that:

The function is increasing on $x < \frac{4}{9}$ and decreasing on $x > \frac{4}{9}$.

Thus, the correct intervals of increase and decrease are:

$\searrow:x>\frac{4}{9}\\\nearrow:x<\frac{4}{9}$

To determine the intervals of increase and decrease for the given function $y = -\left(x + 6\frac{1}{2}\right)^2 - 2\frac{1}{4}$, we need to follow these steps:

• Step 1: Identify the vertex. The vertex form of the parabola is $y = a(x-h)^2 + k$, where $h = -6\frac{1}{2}$ and $k = -2\frac{1}{4}$. The vertex is at the point $(-6\frac{1}{2}, -2\frac{1}{4})$.
• Step 2: Determine the orientation of the parabola. Since $a = -1$, the parabola opens downwards. This implies the function decreases (or falls) from the vertex as $x$ moves away from $-6\frac{1}{2}$.
• Step 3: Define the intervals:
• To the left of the vertex (i.e., for $x < -6\frac{1}{2}$), the function increases because the parabola opens downwards.
• To the right of the vertex (i.e., for $x > -6\frac{1}{2}$), the function decreases.

Therefore, the solution for the intervals is:

$\searrow:x>-6\frac{1}{2}\\\nearrow:x<-6\frac{1}{2}$

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 intervals where the function is increasing or decreasing, follow these steps:

• The given function is $y=\left(x-12\frac{1}{2}\right)^2-4$.
• This is a parabola in the form $y=(x-h)^2 + k$, with vertex $(h, k)$.
• Identify the vertex: $h = 12.5$, $k = -4$.
• The vertex indicates the minima for this parabola since it opens upwards.

For quadratics like this:

• It decreases on the interval to the left of the vertex: $x < 12.5$.
• It increases on the interval to the right of the vertex: $x > 12.5$.

Thus, the function $y=\left(x-12\frac{1}{2}\right)^2-4$ is decreasing for $x < 12.5$ and increasing for $x > 12.5$.

The correct choice, matching this analysis, is choice 4: $\searrow:x < -12\frac{1}{2}\\\nearrow:x > -12\frac{1}{2}$

Therefore, the intervals of increase and decrease are:

$\searrow:x < -12\frac{1}{2}, \nearrow:x > -12\frac{1}{2}$.

To find the intervals of increase and decrease for the function $y = -\left(x+\frac{8}{9}\right)^2$, let's follow these steps:

• Step 1: Identify the vertex of the quadratic function. The vertex form is $y = -\left(x+\frac{8}{9}\right)^2$, where $h = -\frac{8}{9}$. This indicates that the vertex is at $x = -\frac{8}{9}$.
• Step 2: Determine the direction in which the parabola opens. Since the coefficient of the squared term, $a$, is negative ($a = -1$), the parabola opens downwards.
• Step 3: Analyze the behavior of the function around the vertex.

Since the parabola opens downwards:

• The function increases as $x$ approaches the vertex from the left ($x < -\frac{8}{9}$) because the curve slopes downwards towards the vertex.
• The function decreases as $x$ moves away from the vertex to the right ($x > -\frac{8}{9}$) because the curve continues to slope downwards after passing the vertex.

Therefore, the intervals of increase and decrease are:

• Increasing: $x < -\frac{8}{9}$
• Decreasing: $x > -\frac{8}{9}$

The correct answer is choice 2: $\searrow:x > -\frac{8}{9} \\ \nearrow:x < -\frac{8}{9}$

To find the intervals where the function is decreasing, we start by examining the function $y = (x+10)^2 + 6$.

This function is in the vertex form $y = a(x-h)^2 + k$. The vertex $(h, k)$ is located at $(-10, 6)$ since $(x+10)^2$ implies $h = -10$, and the constant $k = 6$ is the y-coordinate of the vertex.

In the vertex form, the parameter $a$ determines the direction in which the parabola opens. In this case, $a = 1$, which is positive, indicating that the parabola opens upwards. For an upward-opening parabola, the function is decreasing in the interval where $x < h$, as the vertex is the lowest point, and the parabola's shape causes it to rise on either side of the vertex.

The vertex is at $x = -10$, so the function $y = (x+10)^2 + 6$ is decreasing for all $x$ less than $-10$.

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

The correct answer to the problem is $x < -10$.

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 $x > -7$.