Vertex Representation: Finding Increasing or Decreasing Domains

To solve this problem, we need to determine the intervals during which the quadratic function $y = (x - 4.6)^2 + 2.1$ is increasing and decreasing.

• Step 1: Identify the vertex of the parabola from the equation, which is presented in vertex form $y = (x - h)^2 + k$. Here, the vertex is $(4.6, 2.1)$.

• Step 2: Determine the direction of the parabola by examining the sign of the coefficient of the squared term. Since $a = 1$ (positive), the parabola opens upwards.

• Step 3: Identify intervals of increase and decrease:

  • For a parabola opening upwards, the function decreases on the interval $x < 4.6$ and increases on the interval $x > 4.6$.

Therefore, the intervals of increase and decrease for the given function are as follows:
Decreasing: $x < 4.6$
Increasing: $x > 4.6$

This matches choice 2 from the given options, confirming that our analysis was correct.

In conclusion, the solution to the problem is:
$\searrow:x<4.6\\\nearrow:x>4.6$

To solve this problem, we'll examine how the function behaves based on the structure given:

1. Rearrange the equation if needed: $2^y = -\left( x + \frac{1}{6} \right)$ implies: $y = \log_2 \left(-\left(x + \frac{1}{6}\right)\right)$. This hints $y$ is undefined unless $x < -\frac{1}{6}$; otherwise, the logarithm argument is non-positive.

2. Recognize: Derived behavior as $x \to -\frac{1}{6}^-$, $y$ shoots toward large negative values (approaches as $-\infty$).

3. Here, solving directly for a derivative doesn't computationally proceed without explicit form, but relies on boundary behavior.

4. Therefore, examine if behavior up to $x = -\frac{1}{6}$ makes it decrease (progressively smaller $y$ as $x$ reduces), and afterwards (impossible), turns - thus indicating:

The function decreases relative to $x > -\frac{1}{6}$

There's no valid interval for $x < -\frac{1}{6}$.

Thus, the solution highlights these ranges:

Therefore, the intervals are given by:

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

Hence, the correct answer choice is:

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

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 solve the problem, let's first identify the form and properties of the given function: $y = (x - 2\frac{1}{9})^2 + \frac{5}{6}$.

The function is a quadratic in vertex form, $y = (x - h)^2 + k$, where $h = 2\frac{1}{9}$ and $k = \frac{5}{6}$. In this form, the vertex is at $x = 2\frac{1}{9}$. The coefficient of the squared term is positive, indicating that the parabola opens upwards.

The vertex point $(2\frac{1}{9}, \frac{5}{6})$ is the minimum point of the parabola. For a quadratic function opening upwards:

• The function decreases when $x < 2\frac{1}{9}$.
• The function increases when $x > 2\frac{1}{9}$.

Therefore, the intervals of decrease and increase are as follows:
- Decreasing: $x < 2\frac{1}{9}$
- Increasing: $x > 2\frac{1}{9}$

Comparing this with the answer choices, the correct choice is:

$\searrow:x<2\frac{1}{9}\\\nearrow:x>2\frac{1}{9}$

To solve this problem, we will analyze the quadratic function given in vertex form.

The function is $y = \left(x - 3\frac{1}{11}\right)^2$, which is in the form $y = (x - h)^2$. Here, $h = 3\frac{1}{11}$, which represents the vertex of the parabola.

For a quadratic function in the vertex form $y = (x - h)^2$:

• The function is decreasing on the interval $x < h$.
• The function is increasing on the interval $x > h$.

Because the function opens upwards (as the coefficient of $(x - h)^2$ is positive), it decreases on the left side of the vertex and increases on its right side.

Therefore, based on the vertex $x = 3\frac{1}{11}$:

Thus, the intervals you are looking for are:

$\searrow:x<3\frac{1}{11}\\\nearrow:x>3\frac{1}{11}$

Comparing this result to the given choices, the correct choice is:

Choice 3: $\searrow:x<3\frac{1}{11}\\\nearrow:x>3\frac{1}{11}$

To solve this problem, we'll determine where the quadratic function $y = (x+8)^2 - 2\frac{1}{4}$ is increasing or decreasing.

This function is in the vertex form: $y = a(x-h)^2 + k$, where $a$ indicates whether the parabola opens upwards ($a > 0$) or downwards ($a < 0$). Here, $a = 1$, indicating the parabola opens upwards.

Let's identify the vertex:

• The vertex form is $(x+8)^2$, thus the vertex $(h, k) = (-8, -2\frac{1}{4})$.

The function is a parabola that opens upwards, so it is decreasing on the left of the vertex and increasing on the right. Specifically:

• Decreasing on the interval $x < -8$.
• Increasing on the interval $x > -8$.

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

Decreasing: $x < -8$

Increasing: $x > -8$

Thus, the correct conclusion for the intervals of increase and decrease is:

$\searrow: x < -8 \\ \nearrow: x > -8$

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 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:
$\searrow:x<8\frac{1}{6}\\\nearrow:x>8\frac{1}{6}$.

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 $y = -\left(x-\frac{4}{9}\right)^2 + 1$, 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 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: $\searrow:x > -\frac{8}{9} \\ \nearrow:x < -\frac{8}{9}$

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:

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

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 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 $x > -10$.

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: $x > 12$.

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 where the function $y = (x+15)^2 + 6$ is decreasing, follow these steps:

• Step 1: Recognize the vertex form of the quadratic $y = (x-h)^2 + k$ where $h = -15$ and $k = 6$. Thus, the vertex is at $x = -15$, $y = 6$.
• Step 2: Understand that this is an upward-opening parabola because the coefficient of the squared term (1) is positive. The function decreases to the left of the vertex.
• Step 3: The decreasing interval is to the left of the vertex $x = -15$. Therefore, the function decreases as $x$ goes from $-\infty$ to $-15$.

Therefore, the function decreases for $x < -15$.

The correct answer choice is: $x < -15$.

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 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 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 $x < 5$.