Find Increasing Intervals for y = (x-9)(5-x): Quadratic Function Analysis

To determine where the function $y = (x-9)(5-x)$ is increasing, we follow these steps:

• Step 1 - Expand the expression: Start by expanding the given function:
  $y = (x-9)(5-x) = x \cdot 5 - x \cdot x - 9 \cdot 5 + 9 \cdot x = 5x - x^2 - 45 + 9x$.
  Simplifying gives: $y = -x^2 + 14x - 45$.

• Step 2 - Differentiate to find $y'$: Compute the first derivative with respect to $x$:
  $y' = \frac{d}{dx} (-x^2 + 14x - 45) = -2x + 14$.

• Step 3 - Solve $y' = 0$ for critical points:
  Set $y'$ equal to zero:
  $-2x + 14 = 0$,
  $2x = 14$,
  $x = 7$.

• Step 4 - Test intervals using the derivative:
  We analyze the sign of the derivative $y' = -2x + 14$ on intervals determined by the critical point $x = 7$.

• For $x < 7$, choose $x = 6$, then $y' = -2(6) + 14 = -12 + 14 = 2$, which is positive. Hence, the function is increasing here.

• For $x > 7$, choose $x = 8$, then $y' = -2(8) + 14 = -16 + 14 = -2$, which is negative. Hence, the function is decreasing here.

Conclusion: The function $y = (x-9)(5-x)$ is increasing on the interval $-\infty < x < 7$.

This matches the correct answer choice (2):

$x<7$