Increasing and Decreasing Domain of a Parabola: Intercept form of quadratic equation

First, we need to express the given function in a form that's easy to differentiate:

The original function is $y = (x-6)(x+6)$. Expanding this, we have:

$y = x^2 - 36$.

Next, we'll find the derivative of this quadratic function to determine the intervals where the function is increasing. The derivative will provide the slope of the tangent at any point on the function:

The derivative of $y = x^2 - 36$ is:

$y' = 2x$.

Now, we determine where the derivative $2x$ is positive. A function is increasing where its derivative is positive:

Solve $2x > 0$:

$x > 0$.

This shows that the function is increasing on the interval where $x > 0$.

Therefore, the solution to the problem is that the function is increasing for $x > 0$.

Let's solve this problem step-by-step:

The function is given by:

$y = (3x + 1)(1 - 3x)$

We can first find the derivative $y'$ to determine where the function is increasing:

• First, expand the product: $y = 3x + 1 - 9x^2 - 3x$
• Simplify: $y = -9x^2 + 0x + 1$

Now the function looks like this quadratic form $y = -9x^2 + 1$.

Next, compute the derivative:

$\frac{dy}{dx} = -18x$

To find the critical points, set $\frac{dy}{dx} = 0$:

$-18x = 0$

Solving, we find $x = 0$.

Now analyze the sign of $\frac{dy}{dx} = -18x$ around this critical point:

• For $x < 0$, $-18x > 0$. Therefore, the function is increasing.
• For $x > 0$, $-18x < 0$. Therefore, the function is decreasing.

Therefore, the solution is that the function is increasing on the interval:

$x < 0$.

The function given is $y = (x-4)(-x+6)$. To analyze its behavior, we first convert this into a standard quadratic form by expanding:

$y = (x-4)(-x+6) = -x^2 + 6x + 4x - 24 = -x^2 + 10x - 24$.

The derivative with respect to $x$ of the function $y = -x^2 + 10x - 24$ is $\frac{dy}{dx} = -2x + 10$.

To find critical points, we set the derivative equal to zero:

$-2x + 10 = 0$

Solving for $x$, we find:

$-2x = -10$

$x = 5$.

Next, we test intervals around the critical point $x = 5$:

• For $x < 5$, choose a test point like $x = 0$: $-2(0) + 10 = 10 > 0$, so the derivative is positive, indicating the function is increasing.
• For $x > 5$, choose a test point like $x = 6$: $-2(6) + 10 = -2 < 0$, so the derivative is negative, indicating the function is decreasing.

Therefore, the function $y = (x-4)(-x+6)$ is decreasing for $x > 5$.

Thus, the correct answer is $x > 5$.

The function $y = (x+1)(x+5)$ is in intercept form, and we can start by expanding it:

$y = x^2 + 6x + 5$.

We take the derivative of the quadratic function with respect to $x$ to find the critical points:

$y' = \frac{d}{dx}(x^2 + 6x + 5) = 2x + 6$.

Set the derivative equal to zero to find any critical points:

$2x + 6 = 0$.
Solving for $x$, we get $2x = -6$ or $x = -3$.

This critical point, $x = -3$, will help us break the number line into intervals to test whether the derivative is positive or negative.

We examine intervals to determine where the function is decreasing by using test points:

• Interval $x < -3$: Choose a test point (e.g., $x = -4$)
  $y'(-4) = 2(-4) + 6 = -8 + 6 = -2$ (negative)
• Interval $x > -3$: Choose a test point (e.g., $x = 0$)
  $y'(0) = 2(0) + 6 = 6$ (positive)

For $x < -3$, $y'$ is negative, indicating the function is decreasing in this interval.

Therefore, the interval where the function is decreasing is $x < -3$.

To determine the decreasing intervals of the function $y = (4x + 8)(-x + 2)$, we follow these steps:

• Step 1: Expand the function.
• Step 2: Compute its derivative.
• Step 3: Find where the derivative is negative.

Step 1: Expand the Function
First, let's expand $y = (4x + 8)(-x + 2)$:

$y = 4x(-x) + 4x(2) + 8(-x) + 8(2)$

$y = -4x^2 + 8x - 8x + 16$

Simplifying, we have $y = -4x^2 + 16$.

Step 2: Compute the Derivative
The derivative of $y$ with respect to $x$ is:

$y' = \frac{d}{dx}(-4x^2 + 16) = -8x$.

Step 3: Find where the Derivative is Negative
We need to solve for $y' < 0$:

$-8x < 0$

This implies $x > 0$.

Therefore, the function $y = (4x + 8)(-x + 2)$ is decreasing on the interval $x > 0$.

To determine the intervals where the function $y = (x+10)(x-8)$ is increasing, we will follow these steps:

• Step 1: Expand the expression for clarity.
• Step 2: Find the derivative $y'$ of the expanded function.
• Step 3: Determine where $y' > 0$ to identify increasing intervals.

Step 1: Expand the function $y = (x+10)(x-8)$.

Expanding gives: $y = x^2 + 2x - 80$.

Step 2: Find the derivative $y'$ of $y = x^2 + 2x - 80$.

$y' = \frac{d}{dx}(x^2 + 2x - 80) = 2x + 2$.

Step 3: Find where the function is increasing by solving $y' > 0$.

$2x + 2 > 0$.

Solve for $x$:

$2x > -2$

$x > -1$.

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

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

To determine where the function $y = (3x+3)(9-x)$ is increasing, we will use the following steps:

• Step 1: Expand and simplify the quadratic expression.
• Step 2: Find the derivative of the function.
• Step 3: Determine where the derivative is positive.

Let's go through these steps:

Step 1: Expand and simplify the quadratic expression:
The function given is $y = (3x + 3)(9 - x)$.

We expand this expression:
$y = 3x \cdot 9 + 3x \cdot (-x) + 3 \cdot 9 + 3 \cdot (-x)$.
This simplifies to:
$y = 27x - 3x^2 + 27 - 3x$.
Combining like terms, we get the quadratic equation:
$y = -3x^2 + 24x + 27$.

Step 2: Find the derivative of the function:
The quadratic equation found is $y = -3x^2 + 24x + 27$.
Taking the derivative, we have:
$y' = \frac{d}{dx}(-3x^2 + 24x + 27) = -6x + 24$.

Step 3: Determine where the derivative is positive:
To find where the function is increasing, solve the inequality:
$y' = -6x + 24 > 0$.
This simplifies to:
$-6x > -24$.
Dividing both sides by -6 (and remembering to reverse the inequality sign) gives:
$x < 4$.

Thus, the function is increasing on the interval where $x < 4$.

Therefore, the solution to the problem is $x < 4$..

To determine where the function $y = (x+1)(7-x)$ is increasing, follow these steps:

• Step 1: Expand the function. $y = (x+1)(7-x) = 7x - x^2 + 7 - x = -x^2 + 6x + 7$
• Step 2: Differentiate the function. $\frac{dy}{dx} = \frac{d}{dx}(-x^2 + 6x + 7) = -2x + 6$
• Step 3: Identify the intervals where the derivative is positive. $-2x + 6 > 0$
• Step 4: Solve the inequality. $-2x + 6 > 0$ $6 > 2x$ $3 > x$ Thus, the function is increasing for $x < 3$.

Therefore, the function $y = (x+1)(7-x)$ is increasing on the interval $x < 3$.

Matching this result with the given choices, the correct choice is:

Choice 3: $x < 3$

To determine where the function $y = (x-4)(x+2)$ is decreasing, we will first convert the product form to standard form:

Step 1: Expand the function: $y = x^2 - 4x + 2x - 8 = x^2 - 2x - 8$

Step 2: Differentiate the function with respect to $x$ to find the derivative $y'$: $y' = \frac{d}{dx}(x^2 - 2x - 8) = 2x - 2$

Step 3: Determine where the derivative is negative: $2x - 2 < 0$

Step 4: Solve for $x$: $2x < 2$ $x < 1$

Therefore, the function $y = (x-4)(x+2)$ is decreasing for $x < 1$.

This corresponds to choice 2: $x<1$

To determine where the function $y = (x-6)(x+6)$ is decreasing, we analyze the quadratic function in its factored form.

Step 1: Identify the roots and the vertex.

• The roots of the function are $x = 6$ and $x = -6$.
• The vertex is exactly halfway between these roots, located at $x = \frac{6 + (-6)}{2} = 0$.

Step 2: Determine the behavior on each side of the vertex.

• This function represents a standard upward-opening parabola because it can be rewritten as $y = x^2 - 36$, which has a positive leading coefficient.
• The parabola decreases as it approaches the vertex from the left and increases as it moves away to the right. Thus, the function is decreasing for values of $x$ less than the vertex, $x = 0$.

Step 3: State the interval where the function is decreasing.

The function is decreasing on the interval $x < 0$.

The correct solution to the problem, where the function is decreasing, is $x < 0$.

To determine where the function $y = (x-9)(5-x)$ is increasing, we follow these steps:

• Step 1 - Expand the expression: Start by expanding the given function:
  $y = (x-9)(5-x) = x \cdot 5 - x \cdot x - 9 \cdot 5 + 9 \cdot x = 5x - x^2 - 45 + 9x$.
  Simplifying gives: $y = -x^2 + 14x - 45$.

• Step 2 - Differentiate to find $y'$: Compute the first derivative with respect to $x$:
  $y' = \frac{d}{dx} (-x^2 + 14x - 45) = -2x + 14$.

• Step 3 - Solve $y' = 0$ for critical points:
  Set $y'$ equal to zero:
  $-2x + 14 = 0$,
  $2x = 14$,
  $x = 7$.

• Step 4 - Test intervals using the derivative:
  We analyze the sign of the derivative $y' = -2x + 14$ on intervals determined by the critical point $x = 7$.

• For $x < 7$, choose $x = 6$, then $y' = -2(6) + 14 = -12 + 14 = 2$, which is positive. Hence, the function is increasing here.

• For $x > 7$, choose $x = 8$, then $y' = -2(8) + 14 = -16 + 14 = -2$, which is negative. Hence, the function is decreasing here.

Conclusion: The function $y = (x-9)(5-x)$ is increasing on the interval $-\infty < x < 7$.

This matches the correct answer choice (2):

$x<7$

To find the intervals on which the quadratic function $y = (x+6)(x-8)$ is increasing, we perform the following steps:

Step 1: Expand the quadratic to standard form:
$y = (x+6)(x-8) = x^2 - 2x - 48$

Step 2: Find the derivative of the function:
The derivative, $f'(x)$, is found by differentiating $x^2 - 2x - 48$:
$f'(x) = 2x - 2$

Step 3: Determine where the derivative is positive:
Set $f'(x) > 0$:
$2x - 2 > 0$

Solve for $x$:
$2x > 2$
$x > 1$

Therefore, the function $y = (x+6)(x-8)$ is increasing on the interval $x > 1$.

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

To find the intervals where the function is decreasing, we'll do the following:

• Step 1: Rewrite the function in standard form.
• Step 2: Determine the direction in which the parabola opens.
• Step 3: Identify the vertex and use it to determine decreasing intervals.

Let's begin by expanding the given function:

Step 1: The function $y = (x-9)(5-x)$ can be expanded to:

$y = -(x - 9)(x - 5) = -(x^2 - 14x + 45)$

This simplifies to:

$y = -x^2 + 14x - 45$

Step 2: Analyze the parabola:

The quadratic equation $-x^2 + 14x - 45$ has a negative leading coefficient (-1), indicating that the parabola opens downward.

Step 3: Find the vertex:

The vertex of a parabola $ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$. Here, $a = -1$ and $b = 14$.

Thus, the x-coordinate of the vertex is $x = -\frac{14}{2(-1)} = 7$.

Because the parabola opens downward, the function decreases after the vertex. Consequently, the function is decreasing for:

$x > 7$

To find the intervals where the function $y = (2x + 10)(3 - x)$ is decreasing, we need to follow a systematic approach:

• Step 1: Expand the function to standard form.
  $y = (2x + 10)(3 - x)$
  $y = 6 - 2x - 30 + 10x$
  $y = -2x^2 + 10x + 6$
• Step 2: Differentiate the function to find its derivative.
  $y' = \frac{d}{dx}(-2x^2 + 10x + 6) = -4x + 10$
• Step 3: Determine critical points by setting the derivative to zero and solving.
  $-4x + 10 = 0$
  $4x = 10$
  $x = \frac{5}{2}$
• Step 4: Identify where the derivative is negative, to find where the function is decreasing.
  Divide the number line into intervals based on this critical point: $x < \frac{5}{2}$ and $x > \frac{5}{2}$.
  Test the intervals in the derivative:
  For $x < \frac{5}{2}$, choose $x = 0$:
  $y'(0) = -4(0) + 10 = 10 > 0$ (positive, so increasing)
  For $x > \frac{5}{2}$, choose $x = 3$:
  $y'(3) = -4(3) + 10 = -12 + 10 = -2 < 0$ (negative, so decreasing)

Therefore, the function $y = (2x + 10)(3 - x)$ is decreasing for: $x > \frac{5}{2}$.

The critical point $\frac{5}{2}$ indicates a turning point from increasing to decreasing.

Thus, the correct choice and solution is: $x > -1$

To find the intervals where the function $y = (x+1)(7-x)$ is decreasing, we begin by expanding the function:
$y = (x+1)(7-x) = 7x - x^2 + 7 - x = -x^2 + 6x + 7$.

The function is now in the form $y = -x^2 + 6x + 7$.
This is a quadratic function, opening downward because the coefficient of $x^2$ is negative.

Let's find the critical points by taking the derivative and setting it to zero.
The derivative of $y$ is$\ y' = \frac{d}{dx}(-x^2 + 6x + 7) = -2x + 6$.
Solving $-2x + 6 = 0$ gives $x = 3$.

The vertex, $x = 3$, is where the function changes from increasing to decreasing.

To determine the interval where the function is decreasing, consider the derivative:$<br> -2x + 6 < 0$.
Solving gives $-2x < -6$, resulting in $x > 3$.

Therefore, the function is decreasing for $x > 3$.

The correct answer is: $x > 3$

To find the intervals where the function $y = (x+1)(x+5)$ is increasing, we need to analyze its derivative.

We start by expanding the function: $y = (x+1)(x+5) = x^2 + 6x + 5$.

Next, we find the derivative: $y' = \frac{d}{dx}(x^2 + 6x + 5) = 2x + 6$.

To find where the function is increasing, solve the inequality $2x + 6 > 0$:

• Subtract 6 from both sides: $2x > -6$
• Divide by 2: $x > -3$

This tells us that the function is increasing on the interval $x > -3$.

By analyzing the derivative, the function transitions at $x = -3$, from decreasing (when $x < -3$) to increasing (when $x > -3$).

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

To find where the function $y = (4x + 8)(-x + 2)$ is increasing, we first need to express it as a standard quadratic function.

First, expand the product:

$y = (4x + 8)(-x + 2)$

$= 4x(-x) + 4x(2) + 8(-x) + 8(2)$

$= -4x^2 + 8x - 8x + 16$

Simplify this to: $y = -4x^2 + 16$

Now, differentiate $y$ with respect to $x$:

$y' = \frac{d}{dx} (-4x^2 + 16)$ = $-8x$

The function is increasing where its derivative is positive:

$-8x > 0$

Solving the inequality, we have:

$x < 0$

Therefore, the function is increasing on the interval $x < 0$.

The correct choice is therefore $x < 0$.

To find where the function $y = (x+10)(x-8)$ is decreasing, let's follow these steps:

• Expand the Function:

First, let's expand the product:

$y = (x+10)(x-8) = x^2 + 2x - 80$

• Differentiate the Function:

Find the derivative $y'$ of the function:

$y' = \frac{d}{dx}(x^2 + 2x - 80) = 2x + 2$

• Find Critical Points:

To find critical points, set the derivative equal to zero:

$2x + 2 = 0$

$2x = -2$

$x = -1$

• Determine Increasing or Decreasing Intervals:

The critical point is $x = -1$. We now analyze the sign of the derivative $y' = 2x + 2$ across the interval:

- For $x < -1$, choose a test point like $x = -2$:

$y' = 2(-2) + 2 = -4 + 2 = -2$ (negative)

- For $x > -1$, choose a test point like $x = 0$:

$y' = 2(0) + 2 = 2$ (positive)

Since the derivative is negative for $x < -1$, the function is decreasing in this interval.

Conclusion: Therefore, the function is decreasing for $x < -1$.

The correct answer from the given choices is:

$x < -1$

To solve for the intervals where the function $y = (3x+1)(1-3x)$ is decreasing, we employ the derivative approach.

Step 1: Simplify the Quadratic Function
Starting with the function:
$y = (3x+1)(1-3x)$
Expand the expression:
$y = 3x \cdot 1 - 3x \cdot 3x + 1 \cdot 1 - 1 \cdot 3x = 3x - 9x^2 + 1 - 3x$
Combine like terms:
$y = -9x^2 + 3x - 3x + 1 = -9x^2 + 1$.

Step 2: Find the Derivative
Differentiate $y = -9x^2 + 1$ with respect to $x$:
$\frac{dy}{dx} = -18x$.

Step 3: Determine Critical Points
Set the derivative equal to zero to find critical points:
$-18x = 0$
Solving gives:
$x = 0$.

Step 4: Sign Analysis of the Derivative
Check the sign of $\frac{dy}{dx} = -18x$ in the intervals determined by the critical points:

• For $x < 0$, choose $x = -1$: $-18(-1) = 18 > 0$, so increasing.
• For $x = 0$, the derivative is zero, indicating a critical point.
• For $x > 0$, choose $x = 1$: $-18(1) = -18 < 0$, so decreasing.

Thus, the function is decreasing in the interval $x > 0$.

Therefore, the correct answer is $x > 0$.

To solve this problem, follow these steps:

• Step 1: **Expand the Function**
  The given function is $y = (3+x)(x-7)$. Expanding it gives: ${y = x^2 - 4x - 21.}$
• Step 2: **Find the Derivative**
  Compute the derivative of $y$: $y' = 2x - 4.$
• Step 3: **Find Critical Points**
  Set the derivative equal to zero and solve: $2x - 4 = 0 \Rightarrow x = 2.$ This is a critical point where the slope changes from negative to positive or vice versa.
• Step 4: **Test Intervals**
  Choose test points around the critical point $x = 2$ to determine the sign of $y'$:
  For $x < 2$, e.g., $x = 0$: $y'(0) = 2(0) - 4 = -4$ (negative).
  For $x > 2$, e.g., $x = 3$: $y'(3) = 2(3) - 4 = 2$ (positive).

Hence, the function is increasing for $x > 2$.

Thus, the function increases in the interval $x > 2$.