Find Intervals of Increase and Decrease for y = -4x² + 28x

To find the intervals where the function $y = -4x^2 + 28x$ increases or decreases, we will proceed with the following steps:

• Step 1: Differentiate the function to find $y'$.
• Step 2: Set $y'$ to zero and solve for $x$ to find critical points.
• Step 3: Use the critical points to define intervals on the x-axis.
• Step 4: Determine the sign of $y'$ in each interval to establish where the function increases or decreases.

Let's execute these steps in detail:

Step 1: Differentiate the function:
The function is given by $y = -4x^2 + 28x$. The derivative is calculated as follows:

$y' = \frac{d}{dx}(-4x^2 + 28x) = -8x + 28$.

Step 2: Find critical points where $y' = 0$:
Solve $-8x + 28 = 0$:

$-8x = -28$
$x = \frac{-28}{-8} = \frac{28}{8} = \frac{7}{2} = 3\frac{1}{2}$.

Step 3: Define intervals using the critical point $x = 3\frac{1}{2}$:
The intervals are $(-\infty, 3\frac{1}{2})$ and $(3\frac{1}{2}, \infty)$.

Step 4: Test the sign of $y'$ in each interval:

• For $x \in (-\infty, 3\frac{1}{2})$, choose a test point, say $x = 0$:
  $y'(0) = -8(0) + 28 = 28$. The derivative is positive, and the function is increasing.
• For $x \in (3\frac{1}{2}, \infty)$, choose a test point, say $x = 4$:
  $y'(4) = -8(4) + 28 = -32 + 28 = -4$. The derivative is negative, and the function is decreasing.

Therefore, the function is increasing on $(-\infty, 3\frac{1}{2})$ and decreasing on $(3\frac{1}{2}, \infty)$.

In conclusion, the intervals of increase and decrease are expressed as follows:

$\searrow~:x>3\frac{1}{2}~~\\\nearrow~:x<3\frac{1}{2}$