Find Intervals of Increase and Decrease for y = (1/4)x² - 3.5x

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:

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