Find Intervals of Increase and Decrease for y = 2x² + 7x - 9

To solve this problem, we'll follow these steps:

• Step 1: Differentiate the quadratic function $y = 2x^2 + 7x - 9$.
• Step 2: Find the critical point by setting the derivative equal to zero.
• Step 3: Determine the intervals of increase and decrease by testing values around the critical point.

Let's begin with the differentiation:

Step 1: Differentiate $y = 2x^2 + 7x - 9$. The derivative is:

$y' = \frac{d}{dx}(2x^2 + 7x - 9) = 4x + 7$.

Step 2: Find the critical points by setting $y' = 0$:

$4x + 7 = 0$

Solving for $x$, we get:

$4x = -7$

$x = -\frac{7}{4} = -1.75$.

Step 3: Determine intervals by testing points around $x = -1.75$:

• For $x < -1.75$, choose a test point like $x = -2$. Evaluating $y'(-2) = 4(-2) + 7 = -8 + 7 = -1$. Since $-1 < 0$, the function is decreasing.
• For $x > -1.75$, choose a test point like $x = 0$. Evaluating $y'(0) = 4(0) + 7 = 7$. Since $7 > 0$, the function is increasing.

Therefore, the function $y = 2x^2 + 7x - 9$ is decreasing in the interval $x < -1.75$ and increasing in the interval $x > -1.75$.

The final intervals of increase and decrease are:

$\nearrow : x > -1\frac{3}{4}$

$\searrow : x < -1\frac{3}{4}$

The correct answer based on provided choices is:

$\searrow ~: x > -1\frac{3}{4} ~~\\ \nearrow ~: x < -1\frac{3}{4}$