Standard Representation: Finding Increasing or Decreasing Domains

Step 1: Differentiate the Function
First, find the derivative of $y$. The derivative $\frac{dy}{dx}$ of $y = -2x^2 - 8x - 10$ is:

$y' = \frac{d}{dx}(-2x^2 - 8x - 10) = -4x - 8$.

Step 2: Find the Critical Point
Set the derivative equal to zero to find the critical point:

$-4x - 8 = 0$.

Solving for $x$, we get:

$-4x = 8$.

$x = -2$.

Step 3: Analyze the Function Around the Critical Point
The critical point $x = -2$ is where the function changes from increasing to decreasing or vice versa. To find the intervals of increase, check the sign of the derivative on either side of $x = -2$. Since the leading coefficient (-2) in the quadratic function is negative, the parabola opens downwards.

Step 4: Establish the Intervals
- For $x < -2$, choose a test point (e.g., $x = -3$) and substitute into the derivative:

$y' = -4(-3) - 8 = 12 - 8 = 4$.

The derivative is positive, indicating that the function is increasing on the interval $x < -2$.

- For $x > -2$, choose a test point (e.g., $x = 0$) and substitute into the derivative:

$y' = -4(0) - 8 = -8$.

The derivative is negative, indicating that the function is decreasing on the interval $x > -2$.

Therefore, the function is increasing in the interval $x < -2$.

$x < -2$

The function given is $y = 3x^2 - 6x + 4$.

First, let's find the derivative of the function, which will help us determine the intervals of increase.

The derivative is given by $f'(x) = \frac{d}{dx}(3x^2 - 6x + 4) = 6x - 6$.

Next, find where the derivative is zero to locate critical points. Solve $6x - 6 = 0$ to get:

$6x = 6$
$x = 1$

The critical point is $x = 1$. This is where the function changes from decreasing to increasing since quadratic functions have one axis of symmetry and $a > 0$: indicating a parabola opening upwards.

To determine the interval of increase, analyze the sign of $f'(x)$:

• For $x > 1$, $6x - 6 > 0$ which implies $f'(x) > 0$, so the function is increasing.
• For $x < 1$, $6x - 6 < 0$ which implies $f'(x) < 0$, so the function is decreasing.

Thus, the domain of increase for the function is when $x > 1$.

The correct answer is therefore $x > 1$.

The function given is $y = -4x^2 - 8x - 12$. To determine where it is decreasing, we first find the vertex:

• The formula for the x-coordinate of the vertex is $x = -\frac{b}{2a}$.
• Substituting the values for $a$ and $b$ yields $x = -\frac{-8}{2 \times -4} = -1$.

Since the coefficient of $x^2$, which is $a = -4$, is negative, the parabola opens downwards. For a downward-opening parabola, the function decreases to the right of the vertex.

Consequently, the interval where the function decreases is $x > -1$.

Therefore, the solution is $x > -1$.

The function provided is $y = -x^2 + 10x - 16$. We want to identify the intervals where this function is decreasing.

To do so, we first find the vertex of the parabola, which will help us determine the regions of increase and decrease. The vertex of a parabola given by $ax^2 + bx + c$ is located at $x = -\frac{b}{2a}$.

For our function, $a = -1$ and $b = 10$. Substituting these values into the formula, we calculate:

$x = -\frac{10}{2 \times -1} = \frac{10}{2} = 5$.

This means the vertex of the parabola is at $x = 5$. Since the coefficient of $x^2$ (i.e., $a$) is negative, the parabola opens downward. Consequently, the function decreases to the right of the vertex.

Thus, the interval where the function is decreasing is when $x > 5$.

Therefore, the solution to the problem is $x > 5$.

To determine the domain over which the quadratic function $y = -x^2 + 2x + 35$ is decreasing, we proceed by identifying the vertex of the parabola.

Given the form $y = ax^2 + bx + c$, we have $a = -1$, $b = 2$, and $c = 35$. The x-coordinate of the vertex can be found using the formula:

Substituting $b = 2$ and $a = -1$ into the formula, we calculate:

The vertex of the parabola occurs at $x = 1$. Since the function is a downward-opening parabola (as indicated by the negative coefficient of $-x^2$), the function decreases for all $x$ values greater than the x-coordinate of the vertex.

Therefore, the domain of decrease for the function is $x > 1$.

This matches the answer choice:

$x > 1$

To determine the intervals where the function $y = x^2 - 8x - 20$ is decreasing, follow these steps:

• Step 1: Find the derivative of the function.
  The derivative of $y = x^2 - 8x - 20$ is $y' = 2x - 8$.
• Step 2: Solve for critical points using the derivative.
  Set the derivative equal to zero: $2x - 8 = 0$.
  Solve for $x$: $2x = 8 \Rightarrow x = 4$.
  This is the vertex of the parabola.
• Step 3: Analyze the sign of the derivative.
  The derivative $y' = 2x - 8$ changes sign at $x = 4$.
  - For $x < 4$, $2x - 8 < 0$ (derivative is negative), meaning the function is decreasing.
  - For $x > 4$, $2x - 8 > 0$ (derivative is positive), meaning the function is increasing.
• Step 4: Determine the interval of decrease.
  The function is decreasing on the interval $x < 4$.

Therefore, the function is decreasing for $x < 4$, which corresponds to the interval $(-∞, 4)$. This matches the correct answer: $x < 4$.

To determine where the function $y = -2x^2 - 12x - 16$ is increasing, we follow these steps:

• Step 1: Find the derivative of the function.
  Starting with $y = -2x^2 - 12x - 16$, the derivative is calculated as:
  $\frac{dy}{dx} = -4x - 12$.
• Step 2: Identify the critical points by setting the derivative to zero and solving for $x$:
  $-4x - 12 = 0$
  $-4x = 12$
  $x = -3$.
• Step 3: Determine the sign of the derivative in the intervals divided by the critical point. The critical point divides the real line into intervals: $(-\infty, -3)$ and $(-3, \infty)$.
• Step 4: Test an $x$-value from each interval to evaluate the sign of $\frac{dy}{dx}$:
  For $x = -4$ in the interval $(-\infty, -3)$,
  $\frac{dy}{dx} = -4(-4) - 12 = 16 - 12 = 4$ (positive).
  For $x = 0$ in the interval $(-3, \infty)$,
  $\frac{dy}{dx} = -4(0) - 12 = -12$ (negative).
• Conclusion: The function $y = -2x^2 - 12x - 16$ is increasing in the interval where the derivative is positive, which is $x < -3$.

Therefore, the solution to the problem is $x < -3$.

To determine where the function $y = 2x^2 - 4x + 5$ is decreasing, we follow these steps:

1. Calculate the Derivative:

The derivative of $y$ with respect to $x$ is given by:

$y' = \frac{d}{dx}(2x^2 - 4x + 5) = 4x - 4$

2. Find Critical Points:

Set the derivative equal to zero and solve for $x$:

$4x - 4 = 0$

$4x = 4$

$x = 1$

3. Analyze the Sign of the Derivative:

To determine where the function is decreasing, examine the sign of $y' = 4x - 4$ around $x = 1$:

• For $x < 1$, choose a test point, such as $x = 0$. Substitute into the derivative:

• $y' = 4(0) - 4 = -4$ (negative), indicating that the function is decreasing.

• For $x > 1$, choose a test point, such as $x = 2$. Substitute into the derivative:

• $y' = 4(2) - 4 = 4$ (positive), indicating that the function is increasing.

Conclusion: The function is decreasing when $x < 1$.

Therefore, the interval where the function $y = 2x^2 - 4x + 5$ is decreasing is:

$x < 1$

To determine where the function $y = 2x^2 - 4x + 5$ is increasing, we must first find its derivative and analyze its behavior.

Step 1: Find the derivative of $y$.
The function given is $y = 2x^2 - 4x + 5$. The first derivative, which represents the slope of the tangent at any point $x$, is found by differentiating:
$y' = \frac{d}{dx}(2x^2 - 4x + 5) = 4x - 4$

Step 2: Set the derivative to greater than zero and solve for $x$.
We set the inequality to determine where the function is increasing:
$4x - 4 > 0$

• Add 4 to both sides to isolate the linear term:
  $4x > 4$
• Divide both sides by 4 to solve for $x$:
  $x > 1$

Thus, the function is increasing for all $x > 1$.

Therefore, the correct answer is $x > 1$.

To determine where the function $y = -2x^2 - 8x - 10$ is decreasing, we need to find the vertex of the parabola. The vertex for a quadratic function in the form $y = ax^2 + bx + c$ occurs at the x-value given by $x = -\frac{b}{2a}$.

Step 1: Identify the coefficients $a = -2$ and $b = -8$.

Step 2: Calculate the vertex x-coordinate:

$x = -\frac{-8}{2(-2)} = -\frac{8}{-4} = 2$.

Step 3: Since $a = -2$ is negative, the parabola opens downwards. Thus, the function decreases for x-values greater than $x = -2$.

Therefore, the interval where the function is decreasing is $x > -2$.

The correct choice is $x>-2$.

To find the domain of increase for the function $y = -x^2 + 2x + 35$, let's determine the vertex first.

• Step 1: Identify coefficients in the quadratic equation. Here, $a = -1$, $b = 2$, and $c = 35$.
• Step 2: Use the vertex formula $x = -\frac{b}{2a}$ to find the x-coordinate of the vertex.

Plug in the values for $b$ and $a$:

The x-coordinate of the vertex is $x = 1$.

Since the coefficient $a$ is negative, this means the parabola opens downwards. A parabola opening downward will increase until it reaches the vertex, then start decreasing.

Therefore, the domain on which the function is increasing is $x < 1$.

Therefore, the solution to the problem is $x < 1$.

To find the domain of increase for the function $y = -x^2 - 8x - 20$, we need to determine the vertex of the parabola, as it will partition the function into increasing and decreasing intervals.

The vertex $x$-coordinate for the quadratic function $y = ax^2 + bx + c$ is given by the formula $x = -\frac{b}{2a}$.

Here, $a = -1$, and $b = -8$.

Substitute these values into the vertex formula:

$x = -\frac{-8}{2 \times -1} = \frac{-8}{-2} = 4$.

Thus, the vertex occurs at $x = -4$.

Since the parabola opens downwards (as $a = -1$ is less than zero), the function is increasing to the left of the vertex. Therefore, the domain of increase is $x < -4$.

Thus, the domain of increase for the function is $x < -4$.

To find the intervals where the function is decreasing, follow these steps:

• Step 1: Differentiate the given function: $y = x^2 + 4x + 5$.
• Step 2: The derivative is $y' = 2x + 4$.
• Step 3: Set the derivative less than zero to find decreasing intervals: $2x + 4 < 0$.
• Step 4: Solve $2x + 4 < 0$ for $x$.

Let's perform the calculations:

The derivative of the function is $y' = 2x + 4$.

Set the inequality: $2x + 4 < 0$.
Subtract 4 from both sides: $2x < -4$.
Divide both sides by 2: $x < -2$.

Therefore, the function $y = x^2 + 4x + 5$ is decreasing for $x < -2$.

The correct answer choice is: $x < -2$.

To find the intervals where the function $y = -3x^2 + 12x - 9$ is decreasing, we begin by calculating the derivative:

The derivative of the function is $y' = \frac{d}{dx}(-3x^2 + 12x - 9) = -6x + 12$.

Next, find the critical point using the vertex formula for the x-coordinate, given by:

$x = -\frac{b}{2a} = -\frac{12}{2(-3)} = 2$.

This critical point $x = 2$ is where the derivative changes sign.

Now, we analyze the sign of the derivative $y' = -6x + 12$:

• For $x < 2$, say $x = 0$: $y' = -6(0) + 12 = 12$ (positive).

• For $x > 2$, say $x = 3$: $y' = -6(3) + 12 = -18 + 12 = -6$ (negative).

Therefore, the function is decreasing on the interval $x > 2$.

To find the intervals where the function $y = -3x^2 + 12x - 9$ is increasing, we first need to determine the vertex of the parabola since it will help us identify the change in behavior from increasing to decreasing.

Step 1: Calculate the derivative of the function.
The function is $y = -3x^2 + 12x - 9$. The derivative is calculated as follows:

$y' = \frac{d}{dx} (-3x^2 + 12x - 9) = -6x + 12$

Step 2: Find the critical points by setting the derivative equal to zero:
$-6x + 12 = 0$

Solve for $x$:
$-6x = -12$
$x = 2$

This critical point $x = 2$ represents the $x$-coordinate of the vertex of the parabola.

Step 3: Determine the sign of the derivative on either side of the critical point $x = 2$.
- For $x < 2$, choose a test point, e.g., $x = 1$:
$y'(1) = -6(1) + 12 = 6 > 0$ (Positive, indicating increasing)

- For $x > 2$, choose a test point, e.g., $x = 3$:
$y'(3) = -6(3) + 12 = -6 < 0$ (Negative, indicating decreasing)

Therefore, the function is increasing on the interval where $x < 2$.

The solution to the problem is $x < 2$.

To find the intervals where the function $y = x^2 + 10x + 16$ is increasing, we follow these steps:

• Step 1: Find the Vertex of the Parabola

The vertex of the parabola $y = ax^2 + bx + c$ is located at $x = -\frac{b}{2a}$. For our function, $a = 1$ and $b = 10$. Thus:

$x = -\frac{b}{2a} = -\frac{10}{2 \times 1} = -5$

• Step 2: Determine the Sign of the Derivative

To find where the function is increasing, compute the derivative of $y$:

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

The function is increasing where $y' > 0$:

$2x + 10 > 0$

Solving the inequality:

$2x > -10$

$x > -5$

• Step 3: Conclusion

The function $y = x^2 + 10x + 16$ is increasing for the interval $x > -5$.

Therefore, the solution to the problem is $x > -5$.

To find the intervals where the function $y = x^2 + 2x - 8$ is increasing, follow these steps:

• Step 1: Find the derivative of the function. The derivative, $y'$, is obtained by differentiating $y = x^2 + 2x - 8$ with respect to $x$.
  $y' = \frac{d}{dx}(x^2 + 2x - 8) = 2x + 2$.
• Step 2: Find critical points by setting the derivative equal to zero:
  $2x + 2 = 0$
  Solve for $x$ to find $x = -1$.
• Step 3: Determine the sign of $y'$ on the intervals determined by this critical point. We have two intervals to consider: $(-\infty, -1)$ and $(-1, \infty)$.
• For $x < -1$, choose a test point, such as $x = -2$:
  $y'(-2) = 2(-2) + 2 = -4 + 2 = -2$ (negative, so $y$ is decreasing).
• For $x > -1$, choose a test point, such as $x = 0$:
  $y'(0) = 2(0) + 2 = 2$ (positive, so $y$ is increasing).

The function is increasing on the interval $x > -1$.

Therefore, the solution to the problem is $x > -1$.

To find where the function $y = 3x^2 - 6x + 4$ is decreasing, we first need to find its derivative. This involves the following steps:

• Step 1: Compute the derivative of the function.

The derivative of the function, using the formula for the derivative of a quadratic function $ax^2 + bx + c$, is given by:

$y' = \frac{d}{dx}(3x^2 - 6x + 4) = 6x - 6$

• Step 2: Set the derivative less than zero to find the decreasing interval.

We set the derivative $6x - 6$ less than zero:

$6x - 6 < 0$

Simplifying the inequality, we get:

$6x < 6$

$x < 1$

• Step 3: Identify the interval for decreasing behavior.

The inequality $x < 1$ indicates the interval where the function is decreasing. Note that this function represents a parabola that opens upwards, and hence, it decreases as $x$ approaches the vertex from the left and increases as $x$ moves right after the vertex.

Therefore, the interval where the function $y = 3x^2 - 6x + 4$ is decreasing is $x < 1$.

To solve the problem and find the intervals of increase for the function $y = x^2 + 4x + 5$, we need to follow these steps:

• Step 1: Calculate the derivative of the function.
• Step 2: Determine where the derivative is greater than zero.
• Step 3: Identify the interval where the function is increasing.

Let's perform each step in detail:

Step 1: Calculate the derivative of the function $y = x^2 + 4x + 5$.
The derivative of $y$ with respect to $x$, denoted as $y'$, is obtained by differentiating each term:
$y' = \frac{d}{dx}(x^2) + \frac{d}{dx}(4x) + \frac{d}{dx}(5)$.
This gives $y' = 2x + 4$.

Step 2: Determine where the derivative is greater than zero.
We solve the inequality $2x + 4 > 0$.
Subtract 4 from both sides: $2x > -4$.
Divide by 2: $x > -2$.

Step 3: Identify the interval where the function is increasing.
The function increases on the interval $x > -2$.

Therefore, the solution to the problem is that the function is increasing for $x > -2$.

To solve the problem of finding where the function $y = x^2 + 10x + 16$ is decreasing, we follow these steps:

• Step 1: Compute the derivative of the function.
• Step 2: Set the derivative less than zero to find the critical points where the function is decreasing.
• Step 3: Solve the inequality and interpret the solution.

First, compute the derivative of $y$.
The function is $y = x^2 + 10x + 16$.
The derivative $\frac{dy}{dx}$ is given by: $\frac{dy}{dx} = 2x + 10$

Next, we determine where the derivative is less than zero, indicating a decrease in the function: $2x + 10 < 0$

Solving the inequality:
Subtract 10 from both sides: $2x < -10$
Divide both sides by 2 to isolate $x$: $x < -5$

This solution suggests that the function is decreasing on the interval $x < -5$.

Therefore, the interval where the function $y = x^2 + 10x + 16$ is decreasing is $x < -5$.