Find Intervals of Increase and Decrease for y = -4x² + 18x

To find the intervals where the function $y = -4x^2 + 18x$ is increasing or decreasing, we need to first find its derivative.

The derivative of the function $y$ with respect to $x$ is:

$y' = \frac{d}{dx}(-4x^2 + 18x) = -8x + 18$.

Next, set the derivative to zero to find the critical points:

$-8x + 18 = 0$.

Solving this equation for $x$, we get:

$-8x = -18$.

$x = \frac{18}{8} = \frac{9}{4} = 2.25$.

This means $x = 2.25$ is a critical point, which corresponds to the vertex of the parabola.

Now, we need to determine the sign of $y'$ on either side of $x = 2.25$ to establish the intervals of increase and decrease.

• For $x < 2.25$, choose a test point such as $x = 0$:
  $y'(0) = -8(0) + 18 = 18$ (positive), indicating the function is increasing.
• For $x > 2.25$, choose a test point such as $x = 3$:
  $y'(3) = -8(3) + 18 = -24 + 18 = -6$ (negative), indicating the function is decreasing.

Therefore, the function is:

Increasing when $x < 2.25$.

Decreasing when $x > 2.25$.

Thus, the solution to the given problem is:

$\nearrow : x < 2\frac{1}{4}$ and $\searrow : x > 2\frac{1}{4}$.