Find Intervals of Increase and Decrease for y = -√11x² - 6

To solve the problem of identifying where the function $y = -\sqrt{11}x^2 - 6$ increases and decreases, we follow these steps:

• Step 1: Analyze the quadratic function in the form $ax^2 + bx + c$.
• Step 2: Identify the values $a = -\sqrt{11}$ and $b = 0$.
• Step 3: Use the vertex formula $x = -\frac{b}{2a}$ to determine the x-coordinate of the vertex.
• Step 4: Determine the intervals by considering the direction in which the parabola opens.

Let's work through each step:

Step 1: The given function is $y = -\sqrt{11}x^2 - 6$, a standard quadratic with $a = -\sqrt{11}$, $b = 0$, and $c = -6$.

Step 2: Observe that $a = -\sqrt{11}$, which is negative. This negative sign means the parabola opens downwards.

Step 3: Calculate the vertex's x-coordinate: $x = -\frac{b}{2a} = -\frac{0}{2(-\sqrt{11})} = 0$.

Step 4: Because the parabola opens downward and the vertex is $(0, -6)$, the function decreases on the interval $x > 0$ and increases on the interval $x < 0$.

Thus, we conclude that the function is:

Decreasing on: $x > 0$

Increasing on: $x < 0$

Therefore, the intervals of increase and decrease for the function are:

$\searrow:x>0\\\nearrow:x<0$