Increasing and Decreasing Domain of a Parabola: Using roots

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:

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

The function $y = 2\sqrt{11}x^2 + 3$ is a quadratic function with $a = 2\sqrt{11}$. Since $a > 0$, the parabola opens upwards.

This implies the parabola decreases on the interval $(-\infty, 0)$ and increases on the interval $(0, \infty)$.

The vertex of this parabola is at $x = 0$ because there is no linear term ($b = 0$). Thus, the parabola is:

$\searrow: x < 0$

$\nearrow: x > 0$

Therefore, the intervals are correctly described by choice 4:
$\searrow: x < 0\\\nearrow: x > 0$

To determine the intervals of increase and decrease for the function $y = \frac{1}{\sqrt{3}}x^2 - 1$, we follow these steps:

• Step 1: Calculate the Derivative

The derivative of the function $y = \frac{1}{\sqrt{3}}x^2 - 1$ is found using basic differentiation rules. Differentiating with respect to $x$, we have:

$y' = \frac{d}{dx}\left(\frac{1}{\sqrt{3}}x^2 - 1\right) = \frac{2}{\sqrt{3}}x$.

• Step 2: Find the Critical Point

To find the critical points, set the derivative $y'$ equal to zero:

$\frac{2}{\sqrt{3}}x = 0$

This gives $x = 0$ as the critical point.

• Step 3: Analyze the Sign of the Derivative

Now, we determine the sign of the derivative on either side of the critical point $x = 0$.

• For $x < 0$:

The derivative $y' = \frac{2}{\sqrt{3}}x$ is negative, thus the function is decreasing.

• For $x > 0$:

The derivative $y' = \frac{2}{\sqrt{3}}x$ is positive, thus the function is increasing.

Conclusion:

The function decreases for $x < 0$ and increases for $x > 0$.

Therefore, the intervals of increase and decrease are:

$\nearrow: x > 0$

$\searrow: x < 0$

The correct answer is:

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

This matches the provided answer in choice option 4.

To determine the intervals of increase and decrease for the quadratic function $y = 4x^2 - 80$, follow this approach:

• First, identify the function's form: $y = 4x^2 - 80$, where $a = 4$, $b = 0$, and $c = -80$.
• Find the vertex $x$-coordinate using the formula $x = -\frac{b}{2a} = -\frac{0}{2 \times 4} = 0$.
• Since $a = 4$ which is positive, the parabola opens upwards.
• With the parabola opening upwards, it decreases before the vertex and increases after the vertex.
• Identify the intervals: The function decreases on the interval $x < 0$ and increases on $x > 0$. The turning point (minimum) occurs precisely at $x = 0$.

Therefore, the function decreases for $x < 0$ and increases for $x > 0$.

The intervals of increase and decrease are $\searrow: x < 0$ and $\nearrow:x > 0$.

To find the intervals where the function $y = 6x^2 - 2$ is increasing or decreasing, follow these steps:

• Step 1: Differentiate the function with respect to $x$.
  The derivative of the function $y = 6x^2 - 2$ is $\frac{dy}{dx} = 12x$.
• Step 2: Find the critical points by setting the derivative equal to zero.
  $12x = 0$
• Step 3: Solve for $x$.
  The solution is $x = 0$.
• Step 4: Determine the intervals around the critical point.
  - For $x < 0$, choose a test point such as $x = -1$. Substituting into the derivative gives $12(-1) = -12$, which is negative, indicating the function is decreasing.
  - For $x > 0$, choose a test point such as $x = 1$. Substituting into the derivative gives $12(1) = 12$, which is positive, indicating the function is increasing.

Therefore, the function is decreasing for $x < 0$ and increasing for $x > 0$.

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

To solve the problem of finding the intervals of increase and decrease for the function $y = \sqrt{5}x^2 + 4$, we will follow these steps:

• Step 1: Identify the function $y = \sqrt{5}x^2 + 4$. Here $a = \sqrt{5}$, so the parabola opens upwards.
• Step 2: Compute the derivative of the function, $y' = \frac{dy}{dx}$, which will help us find the critical points where the slope changes.

Calculating the derivative, we have:

$y = \sqrt{5}x^2 + 4$

$y' = \frac{d}{dx} (\sqrt{5}x^2 + 4) = 2\sqrt{5}x$

Step 3: Find the critical points by setting the derivative equal to zero:

$2\sqrt{5}x = 0$

Solving for $x$ gives $x = 0$.

Step 4: Analyze the sign of $y' = 2\sqrt{5}x$ to determine where the function is increasing or decreasing:

• For $x < 0$, $2\sqrt{5}x < 0$, so the function is decreasing.
• For $x > 0$, $2\sqrt{5}x > 0$, so the function is increasing.

Hence, the function decreases on the interval $x < 0$ and increases on the interval $x > 0$.

Therefore, the solution to the problem is that the function is decreasing for $x > 0$ and increasing for $x < 0$.

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

According to the provided answer choices, the correct choice that matches the intervals we found is choice 3: $\searrow:x > 0$ and $\nearrow:x < 0$.

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:

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

To solve this problem, we follow a methodical approach:

• Step 1: Compute the derivative of the function.
• Step 2: Find the critical points where the derivative is zero.
• Step 3: Determine the sign of the derivative on intervals split by the critical points.
• Step 4: Identify intervals of increase and decrease based on the sign of the derivative.

Now, let's work through each step:

Step 1: Given the function $y = \frac{1}{\sqrt{2}}x^2 - 5$, find its derivative:

Calculating the derivative: $y' = \frac{d}{dx} \left( \frac{1}{\sqrt{2}}x^2 - 5 \right) = \frac{2}{\sqrt{2}}x = \sqrt{2}x$.

Step 2: Set the derivative to zero to find the critical points:

$\sqrt{2}x = 0$ implies $x = 0$.

Step 3: Determine the sign of the derivative on the intervals $x < 0$ and $x > 0$:

• For $x < 0$, $\sqrt{2}x < 0$, so $y' < 0$ (the function is decreasing).
• For $x > 0$, $\sqrt{2}x > 0$, so $y' > 0$ (the function is increasing).

Step 4: Thus, the function is decreasing on the interval $x < 0$ and increasing on the interval $x > 0$.

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

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