Find Intervals of Increase and Decrease for y = (1/9)x² + 1(2/3)x

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:

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