Find Intervals of Increase and Decrease for y = -4x² + x + 3

To determine where the function $y = -4x^2 + x + 3$ is increasing or decreasing, we first calculate its derivative. The function can be written as:

$f(x) = -4x^2 + x + 3$.

The derivative $f'(x)$ is found using the power rule:

$f'(x) = \frac{d}{dx}(-4x^2 + x + 3) = -8x + 1$.

To find the critical points, we set the derivative equal to zero:

$-8x + 1 = 0$.

Solve for $x$:

$-8x = -1$

$x = \frac{1}{8}$.

Now, we test intervals around $x = \frac{1}{8}$ to determine where $f'(x)$ is positive (increasing) or negative (decreasing).

• Choose a test point less than $\frac{1}{8}$, such as $x = 0$:

$f'(0) = -8(0) + 1 = 1$, which is positive.

• Choose a test point greater than $\frac{1}{8}$, such as $x = 1$:

$f'(1) = -8(1) + 1 = -7$, which is negative.

Therefore, the function is increasing on the interval $x < \frac{1}{8}$ and decreasing on the interval $x > \frac{1}{8}$.

Consequently, the intervals of increase and decrease for the function $y = -4x^2 + x + 3$ are expressed as:

$\nearrow: x < \frac{1}{8}$ and $\searrow: x > \frac{1}{8}$.