Find Intervals of Increase and Decrease: Analyzing y = (13/2)x² + 9x + 13.5

To solve this problem, we need to analyze the function $y = \frac{13}{2}x^2 + 9x + 13\frac{1}{2}$ and find where it is increasing or decreasing by using its derivative.

• Step 1: Find the derivative of the function.
  The derivative $y'$ of the function is found using standard differentiation:
  $y' = \frac{d}{dx}\left(\frac{13}{2}x^2 + 9x + 13\frac{1}{2}\right)$
  $y' = 2 \cdot \frac{13}{2}x + 9 = 13x + 9$
• Step 2: Find the critical points by setting the derivative equal to zero.
  $13x + 9 = 0$
  Solving for $x$, we find the critical point:
  $13x = -9$
  $x = -\frac{9}{13}$
• Step 3: Test the intervals around the critical point to find where the function is increasing or decreasing.
  We need to evaluate $y' = 13x + 9$ at points less than and greater than $x = -\frac{9}{13}$:
  - Choose a test point in the interval $x < -\frac{9}{13}$, like $x = -1$:
  $y'(-1) = 13(-1) + 9 = -13 + 9 = -4$ (Negative, so decreasing in this interval)
  - Choose a test point in the interval $x > -\frac{9}{13}$, like $x = 0$:
  $y'(0) = 13(0) + 9 = 9$ (Positive, so increasing in this interval)

Thus, the function is decreasing for $x < -\frac{9}{13}$ and increasing for $x > -\frac{9}{13}$.

Therefore, the intervals of increase and decrease for the function are:
$\searrow:x < -\frac{9}{12}\\\nearrow:x > -\frac{9}{12}$.

The correct multiple-choice answer is Choice 3.