Find Intervals of Increase and Decrease: y = -2x^2 + 10x + 12

To solve this problem, we'll determine the intervals of increase and decrease for the quadratic function $y = -2x^2 + 10x + 12$ using calculus:

• Step 1: Find the derivative of the function.
• Step 2: Identify the critical point by setting the derivative to zero.
• Step 3: Evaluate the sign of the derivative around the critical point to determine the function's behavior on each interval.

Let's proceed with the solution:

Step 1: Find the derivative of the function $y$.

The original function is $y = -2x^2 + 10x + 12$.
The derivative is $y' = \frac{d}{dx}(-2x^2 + 10x + 12) = -4x + 10$.

Step 2: Set the derivative to zero to find the critical point.

Setting $-4x + 10 = 0$, we solve for $x$.

$-4x + 10 = 0$
$-4x = -10$
$x = \frac{10}{4} = 2.5$

Step 3: Determine where the function is increasing or decreasing by evaluating the sign of the derivative before and after $x = 2.5$.

Choose test points: One in each interval $x < 2.5$ and $x > 2.5$.

For $x < 2.5$, test a point like $x = 0$:
$y'(0) = -4(0) + 10 = 10$, which is positive, thus the function is increasing in this interval.

For $x > 2.5$, test a point like $x = 3$:
$y'(3) = -4(3) + 10 = -12 + 10 = -2$, which is negative, thus the function is decreasing in this interval.

Conclusively, the function $y = -2x^2 + 10x + 12$ is increasing for $x < 2.5$ and decreasing for $x > 2.5$.

Therefore, the intervals of increase and decrease are:

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