Increasing and Decreasing Intervals of a Parabola

From the graph we can see that the parabola is a smiling parabola,

meaning that its extreme point is a minimum point.

If we describe it in words, until the extreme point the function decreases,

after the extreme point it increases.

Since we measure the progress using X,

we can say that the function increases whenever X is greater than point C, the extreme point.

Mathematically we can write:

X>C

As we already said, as long as X is greater than C, the function increases.

To determine the intervals of increase and decrease for the function $y = 6x^2 - 15x$, we begin by finding its first derivative.

The first derivative of the function is found as follows:

$y = 6x^2 - 15x$

$y' = \frac{d}{dx}(6x^2 - 15x) = 12x - 15$

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

$12x - 15 = 0$

$12x = 15$

$x = \frac{15}{12} = \frac{5}{4} = 1\frac{1}{4}$

The critical point is $x = 1 \frac{1}{4}$. We need to determine the sign of the derivative on either side of this point to identify the intervals of increase and decrease.

• For $x < 1 \frac{1}{4}$, choose $x = 0$:

$y'(0) = 12(0) - 15 = -15$

Since $y' < 0$, the function is decreasing for $x < 1 \frac{1}{4}$.

• For $x > 1 \frac{1}{4}$, choose $x = 2$:

$y'(2) = 12(2) - 15 = 24 - 15 = 9$

Since $y' > 0$, the function is increasing for $x > 1 \frac{1}{4}$.

Therefore, the function increases for $x > 1 \frac{1}{4}$ and decreases for $x < 1 \frac{1}{4}$.

The correct intervals of increase and decrease for the function $y = 6x^2 - 15x$ are:

$\nearrow: x < 1 \frac{1}{4}$ (Increasing)
$\searrow: x > 1 \frac{1}{4}$ (Decreasing)

To determine the intervals of increase and decrease for the function $y = \frac{1}{3}x^2 + 2\frac{1}{3}x$, we will perform the following steps:

• Step 1: Differentiate the function with respect to $x$.
• Step 2: Set the derivative equal to zero to find the critical points.
• Step 3: Use sign analysis on intervals determined by the critical points to identify where the function is increasing or decreasing.

Let's proceed with the solution:

Step 1: Differentiate $y = \frac{1}{3}x^2 + 2\frac{1}{3}x$.

The derivative $f'(x)$ is given by:

$f'(x) = \frac{d}{dx}\left(\frac{1}{3}x^2\right) + \frac{d}{dx}\left(2\frac{1}{3}x\right)$.

This simplifies to:

$f'(x) = \frac{2}{3}x + 2\frac{1}{3}$.

Converting $2\frac{1}{3}$ to an improper fraction gives $\frac{7}{3}$, hence:

$f'(x) = \frac{2}{3}x + \frac{7}{3}$.

Step 2: Solve $f'(x) = 0$ to find critical points.

Set $\frac{2}{3}x + \frac{7}{3} = 0$.

Multiply through by 3 to eliminate fractions:

$2x + 7 = 0$.

This simplifies to:

$2x = -7$ $\Rightarrow x = -\frac{7}{2}$ or $x = -3.5$.

Step 3: Perform sign analysis around the critical point $x = -3.5$.

• For $x < -3.5$, choose a test point like $x = -4$.

$f'(-4) = \frac{2}{3}(-4) + \frac{7}{3} = -\frac{8}{3} + \frac{7}{3} = -\frac{1}{3}$.

This is negative, indicating the function is decreasing on this interval.

• For $x > -3.5$, choose a test point like $x = 0$.

$f'(0) = \frac{2}{3}(0) + \frac{7}{3} = \frac{7}{3}$.

This is positive, indicating the function is increasing on this interval.

Thus, the function is decreasing for $x < -3.5$ and increasing for $x > -3.5$.

Therefore, the correct answer is: $\searrow: x > -3.5$; $\nearrow: x < -3.5$.

To determine the intervals of increase and decrease for the quadratic function $y = -\frac{1}{6}x^2 + 3\frac{2}{3}x$, follow these steps:

• Step 1: Find the derivative of the function:

The function is given by $y = -\frac{1}{6}x^2 + \frac{11}{3}x$.
The derivative, using power rules, is $\frac{dy}{dx} = -\frac{1}{3}x + \frac{11}{3}$.

• Step 2: Set the derivative equal to zero and solve for $x$ (Critical points):

Set $-\frac{1}{3}x + \frac{11}{3} = 0$.
Solving for $x$, we get $\frac{1}{3}x = \frac{11}{3}$,
Thus, $x = 11$.

• Step 3: Determine the intervals by testing values around the critical point:

For $x < 11$, choose any point like $x = 0$:
$\frac{dy}{dx} = -\frac{1}{3}(0) + \frac{11}{3} = \frac{11}{3} > 0$. So, the function is increasing on $x < 11$.
For $x > 11$, choose any point like $x = 12$:
$\frac{dy}{dx} = -\frac{1}{3}(12) + \frac{11}{3} = -4 + \frac{11}{3} = -\frac{1}{3} < 0$. So, the function is decreasing on $x > 11$.

Thus, the function is increasing on the interval $(-\infty, 11)$ and decreasing on the interval $(11, \infty)$.

Therefore, the intervals of increase and decrease for the function are:
$\nearrow$ for $x < 11$; $\searrow$ for $x > 11$.

To determine the intervals where the function is increasing or decreasing, we first differentiate the function.

Given the function:

$y = \frac{1}{4}x^2 - \frac{7}{2}x$

Calculate the first derivative, $y'$, as follows:

$y' = \frac{d}{dx}\left(\frac{1}{4}x^2 - \frac{7}{2}x\right)$

Applying standard differentiation rules:

$y' = \frac{1}{4} \cdot 2x - \frac{7}{2}$

Simplifying this, we get:

$y' = \frac{1}{2}x - \frac{7}{2}$

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

$\frac{1}{2}x - \frac{7}{2} = 0$

Solving for $x$, we multiply the entire equation by 2 to clear the fractions:

$x - 7 = 0$

$x = 7$

This means that the function has a critical point at $x = 7$.

Evaluate the sign of $y'$ around the critical point to determine the intervals of increase and decrease:

• For $x < 7$, choose a test point like $x = 0$:
• $y'(0) = \frac{1}{2}(0) - \frac{7}{2} = -\frac{7}{2}$ (negative)
• For $x > 7$, choose a test point like $x = 8$:
• $y'(8) = \frac{1}{2}(8) - \frac{7}{2} = \frac{4 - 7}{2} = \frac{1}{2}$ (positive)

Therefore, the function is decreasing on the interval $(-\infty, 7)$ and increasing on the interval $(7, \infty)$.

From these analyses, we conclude:

The correct intervals are:

$\nearrow : x < 7$ (increasing)

$\searrow : x > 7$ (decreasing)

Thus, the correct answer choice is:

$\searrow : x > 7 ~~$.
$\nearrow : x < 7$