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

To determine where the function $y = (x-3)^2$ is increasing, consider the following:

• The function is a parabola that opens upwards, centered at the vertex $x = 3$.
• A parabola of the form $y = (x-p)^2$ is increasing on the interval $x > p$.

This means the function $y = (x-3)^2$ begins to increase after the vertex, which is at $x = 3$.

Thus, the area of increase (or ascending area) for this function is when $x > 3$.

Therefore, the correct answer is $3 < x$.

To solve this problem, we will identify where the parabolic function $y = (x-5)^2$ is decreasing.

• Step 1: Recognize the vertex of the parabola. The equation $y = (x-5)^2$ indicates the vertex is at $(5, 0)$.
• Step 2: Understand the direction of the parabola. This parabola opens upward as it is in the form $y = (x-p)^2$.
• Step 3: Determine the decreasing interval. For an upward-opening parabola, it decreases on the left side of the vertex.

From the vertex form of the parabola, we can conclude that the function decreases when $x < 5$.

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

To solve this problem, we will identify and use the vertex of the quadratic function. Knowing that the function $y = (x+4)^2$ is a parabola, we recognize it is in vertex form. The general form of a parabola is $y = (x-h)^2 + k$, hence the vertex of our function is at $(h, k) = (-4, 0)$.

Next, it's important to understand the behavior of the function around this vertex. Since the parabola opens upwards (as can be seen from the positive coefficient of the squared term), it will decrease as $x$ moves towards negative infinity, reach its minimum value at the vertex, and then increase as $x$ becomes larger than the vertex.

Therefore, the function is decreasing on the interval to the left of the vertex. This describes when $x < -4$.

To conclude, the solution to the problem is that the function is decreasing for $x < -4$.

To solve this problem, we will determine when the function $y = (x+5)^2$ is decreasing by using its derivative:

• Step 1: Differentiate $y = (x+5)^2$ with respect to $x$, yielding $\frac{dy}{dx} = 2(x+5)$.
• Step 2: Set the derivative less than zero: $2(x+5) < 0$.
• Step 3: Simplify the inequality to solve for $x$:
  $x + 5 < 0$ implies $x < -5$.

The function is decreasing for values of $x$ that satisfy $x < -5$.

Therefore, the solution is $x < -5$.

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

• Step 1: Identify the critical points where the parabola is zero.
• Step 2: Solve the inequality to determine where the parabola is above the x-axis.
• Step 3: Use these results to specify the correct domain of x-values.

Step 1: Finding the critical points where the function is zero, solve:
$- (x-3)^2 + 4 = 0$.

Re-organize to find:
$-(x-3)^2 = -4$ simplifies to $(x-3)^2 = 4$.

Step 2: Take the square root of both sides:
$x-3 = \pm2$.

This gives critical points $x = 3 + 2 = 5$ and $x = 3 - 2 = 1$.

Step 3: Since the parabola opens downwards, the function is positive between these roots. Therefore, the interval where the function is positive is:

$1 < x < 5$.

Therefore, the solution to the problem is $1 < x < 5$.

We start by recognizing the function $y = -(x+3)^2$, which is a parabola that opens downwards with its vertex at $(-3, 0)$. In the vertex form of a parabola $y = a(x - h)^2 + k$, the values increase until reaching the vertex if the parabola opens downwards.

The expression $y = -(x+3)^2$ indicates that as $x$ moves towards $-3$ from the left, the function's values increase until reaching the vertex since the parabola is opening downwards.

Therefore, the interval over which the function is increasing corresponds to $x < -3$.

Thus, the ascending area of the parabola described by the function $y = -(x+3)^2$ is for $x < -3$.

To solve this problem, we will use the following steps:

• Step 1: Differentiate the function.
• Step 2: Find where the derivative is greater than zero.
• Step 3: Use the inequality to determine the increasing interval.

Let's proceed with the solution:

Step 1: Differentiate the given function with respect to $x$:

$y = (x+3)^2 + 2x^2$

Differentiate each term:

$\frac{d}{dx}((x+3)^2) = 2(x+3) \cdot 1 = 2(x+3)$

$\frac{d}{dx}(2x^2) = 4x$

Combine the derivatives:

$y' = 2(x+3) + 4x = 2x + 6 + 4x$

$y' = 6x + 6$

Step 2: Find the values of $x$ where the derivative is greater than zero:

$6x + 6 > 0$

Simplify the inequality:

$6x > -6$

$x > -1$

Step 3: The solution to the inequality tells us the interval where the function is increasing:

The function is increasing wherever $x > -1$.

Therefore, the correct answer is $-1 < x$.

To solve this problem, we'll take the following approach step-by-step:

• Step 1: Differentiate the function.
• Step 2: Set the derivative greater than zero and solve.
• Step 3: Determine the correct interval for $x$.

Now, let's work through each step:

Step 1: Differentiate the function
Given $y = -(2x + 6)^2$, apply the chain rule to differentiate:
We first use the substitution $u = 2x + 6$, so $y = -u^2$. The derivative $\frac{dy}{du} = -2u$.
Since $u = 2x + 6$, the derivative $\frac{du}{dx} = 2$. Applying the chain rule, we have:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -2u \cdot 2 = -4(2x + 6) = -8x - 24$

Step 2: Determine where the derivative is positive
A function is increasing when its derivative is positive, so we set the derivative greater than zero:

$-8x - 24 > 0$

Solve for $x$:

$-8x > 24$

$x < -3$

Step 3: Conclusion and multiple-choice validation
The area where the function is increasing corresponds to values of $x$ less than $-3$. According to the choices provided, the correct answer is:

$-3 > x$, which corresponds to choice

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

To solve this problem, follow these steps:

The function given is $y = -(x-6)^2$. This indicates that the parabola opens downwards because of the negative sign.

Step 1: Identify the vertex of the parabola:

The vertex form of a parabola is $y = a(x-h)^2 + k$. Here, the vertex is at $(6, 0)$.

Step 2: Determine the interval where the function is increasing:

Since the parabola opens downwards, the function is increasing as it approaches the vertex from the left.

Thus, the function is increasing for $x < 6$.

Therefore, the correct answer is $x < 6$.

To determine where the function $y = (x+5)^2 + 2x$ is decreasing, we need to follow these steps:

• Step 1: Differentiate the Function
  The function is $y = (x+5)^2 + 2x$. First, we expand and simplify it:
  $y = (x+5)^2 + 2x = x^2 + 10x + 25 + 2x = x^2 + 12x + 25$.
  Now, compute the derivative:
  $y' = \frac{d}{dx} (x^2 + 12x + 25) = 2x + 12$.
• Step 2: Identify Intervals of Decrease
  Set the derivative less than zero to find where the function is decreasing:
  $2x + 12 < 0$.
  Solve for $x$:
  $2x < -12$
  $x < -6$.

Therefore, the function $y = (x+5)^2 + 2x$ is decreasing for $x < -6$.

The correct answer is $x < -6$.