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 find the intervals where the function $y = -4x^2 + 28x$ increases or decreases, we will proceed with the following steps:

• Step 1: Differentiate the function to find $y'$.
• Step 2: Set $y'$ to zero and solve for $x$ to find critical points.
• Step 3: Use the critical points to define intervals on the x-axis.
• Step 4: Determine the sign of $y'$ in each interval to establish where the function increases or decreases.

Let's execute these steps in detail:

Step 1: Differentiate the function:
The function is given by $y = -4x^2 + 28x$. The derivative is calculated as follows:

$y' = \frac{d}{dx}(-4x^2 + 28x) = -8x + 28$.

Step 2: Find critical points where $y' = 0$:
Solve $-8x + 28 = 0$:

$-8x = -28$
$x = \frac{-28}{-8} = \frac{28}{8} = \frac{7}{2} = 3\frac{1}{2}$.

Step 3: Define intervals using the critical point $x = 3\frac{1}{2}$:
The intervals are $(-\infty, 3\frac{1}{2})$ and $(3\frac{1}{2}, \infty)$.

Step 4: Test the sign of $y'$ in each interval:

• For $x \in (-\infty, 3\frac{1}{2})$, choose a test point, say $x = 0$:
  $y'(0) = -8(0) + 28 = 28$. The derivative is positive, and the function is increasing.
• For $x \in (3\frac{1}{2}, \infty)$, choose a test point, say $x = 4$:
  $y'(4) = -8(4) + 28 = -32 + 28 = -4$. The derivative is negative, and the function is decreasing.

Therefore, the function is increasing on $(-\infty, 3\frac{1}{2})$ and decreasing on $(3\frac{1}{2}, \infty)$.

In conclusion, the intervals of increase and decrease are expressed as follows:

$\searrow~:x>3\frac{1}{2}~~\\\nearrow~:x<3\frac{1}{2}$

To find the intervals where the function $y = \frac{1}{9}x^2 + \frac{5}{3}x$ increases or decreases, we first compute its derivative.

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

The derivative is: $y' = \frac{d}{dx} \left(\frac{1}{9}x^2 + \frac{5}{3}x\right) = \frac{2}{9}x + \frac{5}{3}$.

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

$\frac{2}{9}x + \frac{5}{3} = 0$.

Multiplying through by 9 to clear fractions: $2x + 15 = 0$.

Solve for $x$: $x = -\frac{15}{2} = -7.5$.

Step 3: Determine the sign of $y'$ on the intervals determined by the critical point $x = -7.5$.

Test values from each of the intervals $(-\infty, -7.5)$ and $(-7.5, \infty)$.

For $x < -7.5$: Choose $x = -8$. Compute $y'(-8)$:

$y'(-8) = \frac{2}{9}(-8) + \frac{5}{3} = -\frac{16}{9} + \frac{5}{3} = -\frac{16}{9} + \frac{15}{9} = -\frac{1}{9}$; which is negative.

For $x > -7.5$: Choose $x = -7$. Compute $y'(-7)$:

$y'(-7) = \frac{2}{9}(-7) + \frac{5}{3} = -\frac{14}{9} + \frac{15}{9} = \frac{1}{9}$; which is positive.

Therefore, the function decreases on the interval $x > -7.5$ and increases on the interval $x < -7.5$.

The correct interpretation in terms of the choices is:

$\searrow~:x > -7\frac{1}{2}$
$\nearrow~:x < -7\frac{1}{2}$

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 find the intervals where the function $y = -4x^2 + 18x$ is increasing or decreasing, we need to first find its derivative.

The derivative of the function $y$ with respect to $x$ is:

$y' = \frac{d}{dx}(-4x^2 + 18x) = -8x + 18$.

Next, set the derivative to zero to find the critical points:

$-8x + 18 = 0$.

Solving this equation for $x$, we get:

$-8x = -18$.

$x = \frac{18}{8} = \frac{9}{4} = 2.25$.

This means $x = 2.25$ is a critical point, which corresponds to the vertex of the parabola.

Now, we need to determine the sign of $y'$ on either side of $x = 2.25$ to establish the intervals of increase and decrease.

• For $x < 2.25$, choose a test point such as $x = 0$:
  $y'(0) = -8(0) + 18 = 18$ (positive), indicating the function is increasing.
• For $x > 2.25$, choose a test point such as $x = 3$:
  $y'(3) = -8(3) + 18 = -24 + 18 = -6$ (negative), indicating the function is decreasing.

Therefore, the function is:

Increasing when $x < 2.25$.

Decreasing when $x > 2.25$.

Thus, the solution to the given problem is:

$\nearrow : x < 2\frac{1}{4}$ and $\searrow : x > 2\frac{1}{4}$.