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 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 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 find the intervals of increase and decrease for the function $y = \frac{1}{5}x^2 + 1\frac{1}{3}x$, we first calculate the derivative to analyze the behavior of the function.

Step 1: Finding the Derivative
The function is $y = \frac{1}{5}x^2 + \frac{4}{3}x$. To find the derivative, we use the power rule:

$y' = \frac{d}{dx}\left(\frac{1}{5}x^2 + \frac{4}{3}x\right) = \frac{2}{5}x + \frac{4}{3}$.

Step 2: Set the Derivative to Zero
To find critical points, set the derivative equal to zero:

$\frac{2}{5}x + \frac{4}{3} = 0$.

Solving for $x$, we multiply the equation by 15 (to eliminate fractions):

$6x + 20 = 0$.
$6x = -20$.
$x = -\frac{20}{6} = -\frac{10}{3}$.

So, the critical point is at $x = -\frac{10}{3}$, or $x = -3\frac{1}{3}$.

Step 3: Determine the Sign of the Derivative
Test the sign of the derivative on intervals around the critical point $x = -3\frac{1}{3}$:

• For $x < -3\frac{1}{3}$, choose $x = -4$. Then $y' = \frac{2}{5}(-4) + \frac{4}{3} = -\frac{8}{5} + \frac{4}{3}$. Compute to check if negative.
• For $x > -3\frac{1}{3}$, choose $x = 0$. Then $y' = \frac{2}{5}(0) + \frac{4}{3} = \frac{4}{3}$, which is positive.

Conclusion:
The function is decreasing ($\searrow$) for $x > -3\frac{1}{3}$ and increasing ($\nearrow$) for $x < -3\frac{1}{3}$.

Therefore, the correct intervals are:

$\searrow~:x > -3\frac{1}{3}$
$\nearrow~:x < -3\frac{1}{3}$

This matches with choice 4 of the provided options.

To determine where the given function increases or decreases, we follow these steps:

• Step 1: Differentiate the given function. Given the function $y = \frac{1}{2}x^2 + 4\frac{3}{5}x$, we first express $4\frac{3}{5}$ as an improper fraction, which is $\frac{23}{5}$. This makes the function $y = \frac{1}{2}x^2 + \frac{23}{5}x$.
• Step 2: Calculate the derivative of the function. The derivative $y'$ is computed as follows: $y' = \frac{d}{dx}\left(\frac{1}{2}x^2 + \frac{23}{5}x\right) = x + \frac{23}{5}.$
• Step 3: Set the derivative to zero to find critical points. Solve the equation $x + \frac{23}{5} = 0$: $x = -\frac{23}{5}.$
• Step 4: Identify intervals around the critical point. The critical point is $x = -\frac{23}{5}$, corresponding to a point of potential change from increasing to decreasing or vice versa.
• Step 5: Test the intervals determined by the critical points to find the sign of the derivative, determining whether the function is increasing or decreasing in those intervals.

- For $x < -\frac{23}{5}$, choose a test point such as $x = -5$:
$y'(-5) = -5 + \frac{23}{5} = -5 + 4.6 = -0.4$ which is negative, so the function is decreasing ($\searrow$).

- For $x > -\frac{23}{5}$, choose a test point such as $x = 0$:
$y'(0) = 0 + \frac{23}{5} = 4.6$ which is positive, indicating the function is increasing ($\nearrow$).

Therefore, the function decreases for $x < -\frac{23}{5}$ and increases for $x > -\frac{23}{5}$.

Thus, the intervals of increase and decrease for the function are:

$\searrow:x > -4\frac{3}{5}~~\\ \nearrow:x < -4\frac{3}{5}$