Product Representation: Finding Increasing or Decreasing Domains

To solve the problem of finding intervals of increase and decrease for the given function, follow these steps:

• Step 1: Expand the function.
  We start by expanding $y = \left(2x - \frac{1}{2}\right)\left(x - 2\frac{1}{4}\right)$:
  $y = 2x \cdot x - 2x \cdot 2\frac{1}{4} - \frac{1}{2} \cdot x + \frac{1}{2} \cdot 2\frac{1}{4}$.
  Simplifying, we get:
  $y = 2x^2 - \frac{9}{2}x + \frac{1}{2} \cdot \frac{9}{4}$.
  Thus, $y = 2x^2 - \frac{9}{2}x + \frac{9}{8}$.

• Step 2: Differentiate the function.
  Differentiate $y = 2x^2 - \frac{9}{2}x + \frac{9}{8}$ with respect to $x$:
  $\frac{dy}{dx} = 4x - \frac{9}{2}$.

• Step 3: Find the critical points.
  Set the first derivative equal to zero:
  $4x - \frac{9}{2} = 0$.
  Solving for $x$, we get $4x = \frac{9}{2}$, hence $x = \frac{9}{8}$.

• Step 4: Use the first derivative test.
  Evaluate the sign of $\frac{dy}{dx}$ around the critical point $x = \frac{9}{8}$:
  - For $x < \frac{9}{8}$, choose $x = 1$: $\frac{dy}{dx} = 4(1) - \frac{9}{2} = \frac{8}{2} - \frac{9}{2} = -\frac{1}{2}$ (negative).
  - For $x > \frac{9}{8}$, choose $x = 2$: $\frac{dy}{dx} = 4(2) - \frac{9}{2} = \frac{8}{1} - \frac{9}{2} = \frac{7}{2}$ (positive).
  Thus, the function decreases when $x < \frac{9}{8}$ and increases when $x > \frac{9}{8}$.

Conclusively, the intervals of increase and decrease are:

$\searrow: x < 1\frac{1}{4}, \nearrow: x > 1\frac{1}{4}$.

To solve this problem, follow these steps:

• Step 1: Expand the function
  We have $y = (3x + 1)(4x - 2)$. Expanding this, we get:

$y = 3x \cdot 4x + 3x \cdot (-2) + 1 \cdot 4x + 1 \cdot (-2)$

$y = 12x^2 - 6x + 4x - 2$

$y = 12x^2 - 2x - 2$

• Step 2: Find the derivative
  The derivative $y'$ of the function $y = 12x^2 - 2x - 2$ is:

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

$y' = 24x - 2$

• Step 3: Determine critical points
  Set the derivative equal to zero to find critical points:

$24x - 2 = 0$

$24x = 2$

$x = \frac{1}{12}$

• Step 4: Determine intervals of increase and decrease
  Evaluate the sign of $y' = 24x - 2$ on intervals relative to the critical point $x = \frac{1}{12}$:

Test values:

$y'(0) = 24(0) - 2 = -2$ (Negative, so decreasing)

- For $x > \frac{1}{12}$, choose $x = 1$:

$y'(1) = 24(1) - 2 = 22$ (Positive, so increasing)

Therefore, the function is decreasing for $x < \frac{1}{12}$ and increasing for $x > \frac{1}{12}$.

The correct answer is:

$\searrow:x<\frac{1}{12}\\\nearrow:x>\frac{1}{12}$

To determine the intervals of increase and decrease for the function $y = -\left(4x+31\right)^2$, we follow these steps:

• Step 1: Differentiate the function to find its first derivative.
• Step 2: Solve for critical points by setting the derivative equal to zero.
• Step 3: Analyze the sign of the derivative in intervals determined by the critical point.

Now, let's work through each step:

Step 1: Differentiate the function.
The function is $y = -\left(4x+31\right)^2$. Applying the chain rule gives us: $y' = \frac{d}{dx}\left[-(4x+31)^2\right] = -2(4x+31) \cdot \frac{d}{dx}(4x+31) = -2(4x+31) \cdot 4 = -8(4x+31)$ Thus, the derivative is $y' = -8(4x+31)$.

Step 2: Solve for critical points.
Set the derivative equal to zero: $-8(4x+31) = 0$ $4x+31 = 0$ $4x = -31$ $x = -\frac{31}{4} = -7\frac{3}{4}$ This critical point divides the x-axis into two intervals: $x < -7\frac{3}{4}$ and $x > -7\frac{3}{4}$.

Step 3: Analyze the sign of the derivative.
For $x < -7\frac{3}{4}$, say $x = -8$: $y' = -8(4(-8) + 31) = -8(-32 + 31) = -8(-1) = 8$ The derivative is positive, indicating the function is increasing.

For $x > -7\frac{3}{4}$, say $x = 0$: $y' = -8(0 + 31) = -8 \times 31 = -248$ The derivative is negative, indicating the function is decreasing.

Thus, the function is increasing for $x < -7\frac{3}{4}$ and decreasing for $x > -7\frac{3}{4}$.

Therefore, the solution to the problem is $\searrow:x > -7\frac{3}{4} \\\nearrow:x < -7\frac{3}{4}$.

To find the intervals of increase and decrease for the function $y = (5x - 1)^2$, follow these steps:

• Step 1: Differentiate the function $y = (5x - 1)^2$.
• Step 2: Set the derivative equal to zero to find critical points.
• Step 3: Test intervals around the critical point to determine where the function is increasing or decreasing.

Now, let's work through each step.

Step 1: Differentiate the function.
The given function is $y = (5x - 1)^2$. Using the chain rule, the derivative $y'$ is:

Step 2: Find critical points.
Set $y' = 0$:

Solving for $x$, we get:

Step 3: Determine intervals of increase and decrease.
Test intervals around the critical point $x = \frac{1}{5}$.

• For $x < \frac{1}{5}$ (e.g., $x = 0$), $y' = 50(0) - 10 = -10$. Therefore, $y' < 0$, and the function is decreasing.
• For $x > \frac{1}{5}$ (e.g., $x = 1$), $y' = 50(1) - 10 = 40$. Therefore, $y' > 0$, and the function is increasing.

Thus, the function is decreasing on the interval $x < \frac{1}{5}$ and increasing on the interval $x > \frac{1}{5}$.

Therefore, the correct choice is:

$\searrow: x < \frac{1}{5} \\ \nearrow: x > \frac{1}{5}$

To find where the function $y = \left(\frac{1}{3}x + \frac{1}{7}\right)^2$ is increasing or decreasing, follow these steps:

• Step 1: Calculate the derivative of the function, $y = \left(\frac{1}{3}x + \frac{1}{7}\right)^2$. Use the chain rule: differentiate the outer function and multiply by the derivative of the inner function.
• Step 2: Differentiate: $\frac{d}{dx}\left( \left(\frac{1}{3}x + \frac{1}{7}\right)^2 \right) = 2\left(\frac{1}{3}x + \frac{1}{7}\right) \times \frac{1}{3} = \frac{2}{3}\left(\frac{1}{3}x + \frac{1}{7}\right)$.
• Step 3: Set the derivative equal to zero to find critical points: $\frac{2}{3}\left(\frac{1}{3}x + \frac{1}{7}\right) = 0$.
• Step 4: Solve for $x$: $\frac{1}{3}x + \frac{1}{7} = 0$ implies $\frac{1}{3}x = -\frac{1}{7}$, leading to $x = -\frac{3}{7}$.
• Step 5: Apply the first derivative test around $x = -\frac{3}{7}$: Check sign changes of derivative.
• Step 6: For $x < -\frac{3}{7}$, choose a test point (e.g., $x = -1$), plug into $\frac{2}{3}\left(\frac{1}{3}x + \frac{1}{7}\right)$ and observe negative result indicates decreasing.
• Step 7: For $x > -\frac{3}{7}$, choose another test point (e.g., $x = 0$), plug in and positive result implies increasing.

Therefore, the function $y = \left(\frac{1}{3}x + \frac{1}{7}\right)^2$ decreases for $x < -\frac{3}{7}$ and increases for $x > -\frac{3}{7}$.

Final Solution:

$\searrow:x<-3/7 \\ \nearrow:x>-3/7$, and checking its equivalence in the given answer options as fractions reveals it is choice 1: $\searrow:x<\frac{9}{21} \\ \nearrow:x>\frac{9}{21}$.

To find the intervals of increase and decrease for the function $y = \left(x + \frac{1}{6}\right)\left(-x - 4\frac{1}{9}\right)$, we follow these steps:

• Step 1: Expand the function. This means multiplying out the terms:
  $y = (x + \frac{1}{6})(-x - \frac{37}{9}) = -x^2 - \frac{37}{9}x - \frac{1}{6}x - \frac{37}{54}$
  Simplifying gives $y = -x^2 - \frac{223}{54}x - \frac{37}{54}$.
• Step 2: Find the derivative of the function.
  $\frac{dy}{dx} = -2x - \frac{223}{54}$.
• Step 3: Find critical points by setting the derivative to zero:
  $-2x - \frac{223}{54} = 0$. Solving for $x$ gives $x = -\frac{223}{108}$, which simplifies to $x = -\frac{77}{36}$.
• Step 4: Determine intervals of increase and decrease by testing values around the critical point $x = -\frac{77}{36}$.
  If $x \lt -\frac{77}{36}$, the derivative $-2x - \frac{223}{54}$ is positive, indicating an increasing interval.
  If $x \gt -\frac{77}{36}$, the derivative $-2x - \frac{223}{54}$ is negative, indicating a decreasing interval.

Thus, the function is increasing on the interval $x \lt -\frac{77}{36}$ and decreasing on the interval $x \gt -\frac{77}{36}$.

Therefore, the intervals of increase and decrease are:
$\searrow:x \gt -\frac{77}{36} \\ \nearrow:x \lt -\frac{77}{36}$

The function given is $y = \left(x - \frac{3}{4}\right)^2$. This is a quadratic function with its vertex (or minimum point) at $x = \frac{3}{4}$.

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

• Step 1: Differentiate $y$ with respect to $x$.

$y' = \frac{d}{dx}\left[\left(x - \frac{3}{4}\right)^2\right] = 2\left(x - \frac{3}{4}\right)$

• Step 2: Find the critical points by setting the derivative equal to zero.

$2\left(x - \frac{3}{4}\right) = 0 \quad \Rightarrow \quad x = \frac{3}{4}$

• Step 3: Determine the sign of $y'$ in the intervals divided by the critical point.

For $x < \frac{3}{4}$, $\left(x - \frac{3}{4}\right) < 0$ and hence $y' < 0$, indicating decreasing behavior.

For $x > \frac{3}{4}$, $\left(x - \frac{3}{4}\right) > 0$ and hence $y' > 0$, indicating increasing behavior.

Thus, the function decreases on the interval $(-\infty, \frac{3}{4})$, and increases on the interval $(\frac{3}{4}, \infty)$.

Consequently, the intervals of increase and decrease are:

$\searrow: x > \frac{3}{4} \\ \nearrow: x < \frac{3}{4}$

To solve this problem, let's follow through the steps:

• Step 1: Differentiate the function $y=(2x-16)^2$ with respect to $x$.

The derivative of $y$ is found using the chain rule. The outer function is $u^2$ with $u = (2x-16)$. Thus, the derivative is:

$y' = 2(2x-16) \cdot \frac{d}{dx}(2x-16)$

This simplifies to:

$y' = 2(2x-16) \cdot 2 = 4(2x-16) = 8x - 64$

• Step 2: Find the critical points by setting the derivative to zero.

Set $8x - 64 = 0$:

$8x = 64$

$x = 8$, so $x = 8$ is a critical point.

• Step 3: Analyze each interval determined by this critical point to determine where $y$ is increasing or decreasing.

The number line is split into two intervals by the critical point $x = 8$: $(-∞, 8)$ and $(8, ∞)$.

For $x < 8$ (e.g., $x = 7$):

Evaluate $y'(7) = 8(7) - 64 = 56 - 64 = -8$, so $y'(x) < 0$. Thus, $y$ is decreasing for $x < 8$.

For $x > 8$ (e.g., $x = 9$):

Evaluate $y'(9) = 8(9) - 64 = 72 - 64 = 8$, so $y'(x) > 0$. Thus, $y$ is increasing for $x > 8$.

Therefore, the function $y = (2x-16)^2$ is:
Decreasing on the interval $(-∞, 8)$ and Increasing on the interval $(8, ∞)$.

This corresponds to the correct answer choice:

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

To determine where the function $y = (4x + 16)^2$ is increasing or decreasing, let's analyze its derivative:

Step 1: Differentiate the function.

The function $y = (4x + 16)^2$ is of the form $(u(x))^2$ where $u(x) = 4x + 16$. The derivative of $y$ with respect to $x$ is:

$y' = 2u(x) \cdot u'(x)$

Here, $u'(x) = 4$, so the derivative is:

$y' = 2(4x + 16) \cdot 4 = 8(4x + 16) = 32x + 128$.

Step 2: Find the critical points.

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

$32x + 128 = 0$

$32x = -128$

$x = -4$.

Step 3: Determine the sign of the derivative around the critical point $x = -4$.

• Choose a test point less than $-4$, for example $x = -5$. Then, $y' = 32(-5) + 128 = -32$, which is negative, indicating the function is decreasing.
• Choose a test point greater than $-4$, for example $x = 0$. Then, $y' = 32(0) + 128 = 128$, which is positive, indicating the function is increasing.

Therefore, the function is decreasing on the interval $x < -4$ and increasing on the interval $x > -4$.

The correct answer choice matches these findings:

$\searrow:x<-4\\\nearrow:x>-4$

To solve this problem, we'll begin by finding the derivative of the given function $y = (4x + 22)^2$ with respect to $x$.

The function can be expanded as:

$y = (4x + 22)^2 = 16x^2 + 2 \cdot 4 \cdot 22 \cdot x + 22^2 = 16x^2 + 176x + 484$

Next, find $\frac{dy}{dx}$ by differentiating:

$\frac{dy}{dx} = \frac{d}{dx}(16x^2 + 176x + 484) = 32x + 176$

Set the derivative equal to zero to find the critical point:

$32x + 176 = 0$

$32x = -176$

$x = -\frac{176}{32}$

$x = -\frac{11}{2}$

$x = -5.5$

This critical point, $x = -5.5$, will be the vertex of the parabola, determining where the function changes from decreasing to increasing.

Now, test the sign of $\frac{dy}{dx} = 32x + 176$ in intervals around the critical point:

• For $x < -5.5$, choose $x = -6$:

$32(-6) + 176 = -192 + 176 = -16$

Since the derivative is negative, the function is decreasing.

• For $x > -5.5$, choose $x = -5$:

$32(-5) + 176 = -160 + 176 = 16$

Since the derivative is positive, the function is increasing.

Therefore, the intervals of increase and decrease are:

$\searrow:x < -5\frac{1}{2}$

$\nearrow:x > -5\frac{1}{2}$

Thus, the correct choice from the given options is:

$\searrow:x < -5\frac{1}{2}$

$\nearrow:x > -5\frac{1}{2}$

To determine the intervals of increase and decrease, follow these steps:

• Step 1: Simplify the function. The given function is $y = -(4x + 32)^2$, a quadratic in terms of $x$.
• Step 2: Recognize that this is a downward-facing parabola because the coefficient of $(4x + 32)^2$ is negative.
• Step 3: Find the vertex or the critical point of the parabola. The function is in the form $y = -a(x + h)^2 + k$. Here, $a = 4$, $h = -8$, and $k = 0$.
• Step 4: The vertex occurs at $x = -8$. The function is symmetrical about this point.
• Step 5: Since it's a downwards-opening parabola, the function increases on the left of the vertex and decreases on the right of the vertex.
• Step 6: Thus, the function decreases ($\searrow$) for $x < -8$ and increases ($\nearrow$) for $x > -8$.

Consequently, the intervals of increase and decrease are:
Decreasing: $x < -8$
Increasing: $x > -8$

Therefore, $\searrow: x<-8 \\ \nearrow: x>-8$ is the correct answer.

To find the intervals of increase and decrease for the function $y = (5x - 1)\left(4x - \frac{1}{4}\right)$, we perform the following steps:

• Step 1: Differentiate the function using the product rule.
• Step 2: Find the critical points by setting the derivative to zero.
• Step 3: Determine the intervals where the derivative is positive or negative to infer increasing and decreasing behavior.

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

Step 1: Differentiate the function.
Using the product rule, consider $u = 5x - 1$ and $v = 4x - \frac{1}{4}$. The derivative of the function is:

$y' = \frac{d}{dx}\left[(5x - 1)\left(4x - \frac{1}{4}\right)\right] = (5x - 1)\frac{d}{dx}\left(4x - \frac{1}{4}\right) + \left(4x - \frac{1}{4}\right)\frac{d}{dx}(5x - 1)$

$= (5x - 1) \cdot 4 + \left(4x - \frac{1}{4}\right) \cdot 5 = 20x - 4 + 20x - \frac{5}{4}$

$= 40x - \frac{21}{4}$

Step 2: Find the critical points.
Set the derivative to zero:

$40x - \frac{21}{4} = 0$

Solving for $x$, multiply both sides by 4 to clear the fraction:

$160x - 21 = 0$

$x = \frac{21}{160}$

Step 3: Analyze the sign of the derivative around the critical point to determine increasing or decreasing intervals.
Choose a test point in each interval defined by the critical point $x = \frac{21}{160}$.

• For $x < \frac{21}{160}$, choose $x = 0$ and check the sign of $y' = 40(0) - \frac{21}{4} = -\frac{21}{4}$ which is negative, indicating $y$ is decreasing.
• For $x > \frac{21}{160}$, choose $x = 1$ and check the sign of $y' = 40(1) - \frac{21}{4} = \frac{139}{4}$ which is positive, indicating $y$ is increasing.

Thus, the function decreases for $x < \frac{21}{160}$ and increases for $x > \frac{21}{160}$.

Therefore, the intervals are $\searrow:x<\frac{21}{160}$ and $\nearrow:x>\frac{21}{160}$.

The correct choice is:

$\searrow:x<\frac{21}{160}\\\nearrow:x>\frac{21}{160}$

To solve this problem, we'll find the intervals of increase and decrease for the function $y = (7x + 3)(5x - 2)$.

First, let's expand the function:

$y = (7x + 3)(5x - 2) = 35x^2 + 15x - 14x - 6 = 35x^2 + x - 6$.

Next, compute the derivative $y'$:

$y' = \frac{d}{dx}(35x^2 + x - 6) = 70x + 1$.

To find critical points, set $y' = 0$:

$70x + 1 = 0$

$70x = -1$

$x = -\frac{1}{70}$.

Now, we need to determine the sign of $y'$ in the intervals around the critical point $x = -\frac{1}{70}$:

• For $x < -\frac{1}{70}$, choose a test point such as $x = -1$:
  $y' = 70(-1) + 1 = -70 + 1 = -69$, so $y' < 0$, indicating that the function is decreasing in this interval.
• For $x > -\frac{1}{70}$, choose a test point such as $x = 0$:
  $y' = 70(0) + 1 = 1$, so $y' > 0$, indicating that the function is increasing in this interval.

Putting it all together, we have:

The function is decreasing on the interval $x < -\frac{1}{70}$ and increasing on the interval $x > -\frac{1}{70}$.

Therefore, the intervals of increase and decrease are:

$\searrow:x < -\frac{1}{70} \\ \nearrow:x > -\frac{1}{70}$.

To determine where the function $y=\left(\frac{1}{3}x+\frac{1}{2}\right)^2$ is increasing or decreasing, we need to first find its derivative.

Let's compute the derivative $y'$ of the function:

$y = \left(\frac{1}{3}x + \frac{1}{2}\right)^2$.

Using the chain rule, let $u = \frac{1}{3}x + \frac{1}{2}$, then $y = u^2$.

The derivative of $u^2$ with respect to $x$ is $2u \cdot \frac{du}{dx}$.

Now, find $\frac{du}{dx}$:

$\frac{du}{dx} = \frac{1}{3}$.

Thus, the derivative of the function is:

$y' = 2 \left( \frac{1}{3}x + \frac{1}{2} \right) \cdot \frac{1}{3} = \frac{2}{3} \left( \frac{1}{3}x + \frac{1}{2} \right)$.

Set this derivative to zero to find critical points:

$\frac{2}{3} \left( \frac{1}{3}x + \frac{1}{2} \right) = 0$.

Simplify to find $x$:

$\frac{1}{3}x + \frac{1}{2} = 0$.

$\frac{1}{3}x = -\frac{1}{2}$.

$x = -\frac{1}{2} \times 3$.

$x = -\frac{3}{2}$.

The critical point is at $x = -\frac{3}{2}$.

To determine the nature of intervals around this critical point, test $y'$ on intervals around $x = -\frac{3}{2}$.

- For $x < -\frac{3}{2}$: Choose $x = -2$.
$y' = \frac{2}{3} \left( \frac{1}{3}(-2) + \frac{1}{2} \right) = \frac{2}{3}(-\frac{2}{3} + \frac{1}{2}) < 0$.
$y'$ is negative, so $y$ is decreasing.

- For $x > -\frac{3}{2}$: Choose $x = 0$.
$y' = \frac{2}{3} \left( \frac{1}{3}(0) + \frac{1}{2} \right) = \frac{2}{3} \times \frac{1}{2} > 0$.
$y'$ is positive, so $y$ is increasing.

Thus, the function decreases on $x < -\frac{3}{2}$ and increases on $x > -\frac{3}{2}$.

The intervals of increase and decrease are $\searrow: x > -1\frac{1}{2}$ and $\nearrow: x < -1\frac{1}{2}$.

Analyzing the multiple-choice answers, the correct one matches choice 3.

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

• Step 1: Identify a simpler form of the equation and its derivative.
• Step 2: Find critical points where the derivative is zero.
• Step 3: Determine signs of the derivative on intervals around critical points.

Let's proceed with these steps:

Step 1: Our function is already in vertex form: $y = \left(2x - 2\frac{1}{4}\right)^2$. It's important to note that this term can be simplified as $y = (2x - 2.25)^2$, identifying the vertex at $x = 1.125$, which is $x = 1\frac{1}{8}$.

Step 2: Differentiate the function with respect to $x$. For $y = (2x - 2.25)^2$, use the chain rule:
$y' = 2(2x - 2.25) \cdot 2 = 4(2x - 2.25) = 8x - 9$.

Step 3: Set the derivative to zero to find critical points:
$8x - 9 = 0$ leads to $x = \frac{9}{8} = 1.125$ or $x = 1\frac{1}{8}$.

The function decreases to the left of this point and increases to the right. Specifically:

• If $x < 1\frac{1}{8}$, $8x - 9 < 0$, so $y$ is decreasing.
• If $x > 1\frac{1}{8}$, $8x - 9 > 0$, so $y$ is increasing.

Therefore, the intervals of increase and decrease are:

$\nearrow:x<1\frac{1}{8}$ (Increasing: to the left of the vertex),

$\searrow:x>1\frac{1}{8}$ (Decreasing: to the right of the vertex).

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

• Step 1: Expand the original function
• Step 2: Differentiate the function
• Step 3: Find the critical points by setting the derivative to zero
• Step 4: Determine intervals of increase and decrease based on the derivative's sign

Let's explore each step in detail:
Step 1: Expand the function:

The function $y = (x+1)(x+\frac{1}{6})$ expands to:

$y = x^2 + \frac{1}{6}x + x + \frac{1}{6} = x^2 + \frac{7}{6}x + \frac{1}{6}$.

Step 2: Differentiate the function:
The derivative of the quadratic function $y = x^2 + \frac{7}{6}x + \frac{1}{6}$ is:

$y' = 2x + \frac{7}{6}$.

Step 3: Set the derivative equal to zero to find critical points:
Solve $2x + \frac{7}{6} = 0$.

Subtract $\frac{7}{6}$ from both sides: $2x = -\frac{7}{6}$.

Divide both sides by 2 to solve for $x$:

$x = -\frac{7}{12}$.

Step 4: Determine the sign of the derivative on either side of the critical point:

- For $x < -\frac{7}{12}$, the derivative $y' = 2x + \frac{7}{6}$ is negative, indicating the function is decreasing.

- For $x > -\frac{7}{12}$, the derivative $y'$ is positive, indicating the function is increasing.

Thus, the function is decreasing on $x < -\frac{7}{12}$ and increasing on $x > -\frac{7}{12}$.

Therefore, the final solution is:

$\searrow:x < -\frac{7}{12} \\ \nearrow: x > -\frac{7}{12}$.

Let's solve the problem step by step:

• **Step 1:** Identify the points $p$ and $q$ from the function $y = (x-4.4)(x-2.3)$. Here, $p = 4.4$ and $q = 2.3$.
• **Step 2:** Calculate the vertex $x$-coordinate using the formula for the midpoint: $x = \frac{p+q}{2} = \frac{4.4 + 2.3}{2}$.
• **Step 3:** Compute the value: $x = \frac{6.7}{2} = 3.35$.
• **Step 4:** Since the quadratic has a positive leading coefficient after expansion (implying it opens upwards), the function decreases on $(-\infty, 3.35)$ and increases on $(3.35, \infty)$.

Therefore, the function is decreasing for $x < 3.35$ and increasing for $x > 3.35$.