Increasing and Decreasing Domain of a Parabola: Representation of value a only

Let's solve the problem:

The given function is $y = 3\sqrt{5}x^2$. This is a quadratic function in standard form with $a = 3\sqrt{5}$, which is positive. Quadratics with positive coefficients open upwards and have a distinctive symmetry:

• The interval of decrease: $x < 0$ because the function decreases as we move left from the vertex at the origin.
• The interval of increase: $x > 0$ because the function increases as we move right from the vertex at the origin.

This matches the mathematical property of parabolas where $a > 0$: they decrease until the vertex and then increase past the vertex.

Thus, the intervals of increase and decrease for the function $y = 3\sqrt{5}x^2$ are:

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

The function is given by $y = -4\sqrt{7}x^2$, which is a downward-opening parabola because the coefficient $a = -4\sqrt{7}$ is negative.

The vertex of the parabola is at the origin (0,0). A downward-opening parabola decreases as $x$ moves away from the vertex in the positive $x$-direction, and increases as $x$ moves away from the vertex in the negative $x$-direction.

Thus, the intervals of increase and decrease for this function are:

• The function increases on the interval $(-\infty, 0)$.
• The function decreases on the interval $(0, \infty)$.

Therefore, the intervals of increase and decrease can be denoted as:

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

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

• Simplify the function expression
• Find the derivative of the function
• Determine where the derivative is positive, zero, or negative
• Draw conclusions about the function's increase or decrease intervals

Let's go through each step:

Simplify the Function:
The function given is $y = \sqrt{2x^2}$. We can simplify this to $y = \sqrt{2}|x|$ because $\sqrt{2x^2} = \sqrt{2} \cdot |x|$.

Calculate the Derivative:
Since $y = \sqrt{2}|x|$, let's consider the derivative of $y = \sqrt{2}|x|$. Note that when differentiating absolute value functions: - For $x > 0$, $|x| = x$, so $y = \sqrt{2}x$ and its derivative is $y' = \sqrt{2}$. - For $x < 0$, $|x| = -x$, so $y = -\sqrt{2}x$ and its derivative is $y' = -\sqrt{2}$.

Determine Sign of the Derivative:
Analyzing $y'$: - For $x > 0$, the derivative $y' = \sqrt{2} > 0$, implying the function is increasing on this interval. - For $x < 0$, the derivative $y' = -\sqrt{2} < 0$, implying the function is decreasing on this interval.

Conclusion:
The function decreases on the interval $x < 0$ and increases on the interval $x > 0$.

Thus, the intervals of increase and decrease of the function are $\searrow:x < 0$ (decreasing), and $\nearrow:x > 0$ (increasing).

To solve this problem, we will first differentiate the function $y = \sqrt{3x^2}$. This function can be rewritten as $y = \sqrt{3} \cdot |x|$. Therefore, we differentiate it using the absolute value properties.

Thus, the derivative $y'$ can be written as follows:

$y' = \begin{cases} \sqrt{3} & \text{if } x > 0 \\ -\sqrt{3} & \text{if } x < 0 \end{cases}$

The function has a derivative of $\sqrt{3}$ for $x > 0$ which indicates that the function is decreasing, since the derivative is positive but the graph of absolute value decreases.

Conversely, for $x < 0$, the derivative is $-\sqrt{3}$, indicating the function is increasing since the sign change relates to absolute value properties.

Thus, we determine the function's intervals:

• The function is decreasing on the interval $(0, \infty)$.
• The function is increasing on the interval $(-\infty, 0)$.

Thus, the correct choice is: $\searrow: x > 0 \\ \nearrow: x < 0$

Therefore, the solution is: $\searrow: x > 0 \\ \nearrow: x < 0$.