91Ó°ÊÓ

To find the y-intercepts of a curve, you need to look at where the curve crosses the y-axis. This always happens when the value of x is zero. So, the first step involves substituting zero for x in the given curve equation. In this case, replacing x with zero in the equation \(2x - y + y^3 = 0\) simplifies to \(-y + y^3 = 0\).

From this simplified equation, you can factor out y, resulting in \(y(1 - y^2) = 0\). Solving this equation gives you three potential values of y: 0, 1, and -1. Each of these values represents a point where the curve intercepts the y-axis, giving you the y-intercept points of (0, 0), (0, 1), and (0, -1). Understanding this concept helps us see how the curve behaves as it crosses the y-axis.

**Key Points to Remember:**

• Y-intercepts occur where \(x = 0\).
• Plugging in \(x = 0\) simplifies our equation.
• Factor and solve the equation to find all y-values.

To verify the expression for the derivative of a curve, especially when given implicitly, we must use implicit differentiation. This technique lets us find derivatives for equations not easily solved for y in terms of x. For the given equation \(2x - y + y^3 = 0\), we differentiate each term with respect to x.

We know that the derivative of a constant times x is the constant itself, and since \(x\) is implicitly \(\frac{dx}{dx} = 1\), the derivative of \(2x\) is simply 2. The derivative of \(y\) with respect to x is \(\frac{dy}{dx}\) due to the chain rule, and for \(y^3\), it becomes \(3y^2\frac{dy}{dx}\). Together, these differentiated parts lead to:

\[2 - \frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0\]

Your goal is to isolate \(\frac{dy}{dx}\). Rearrange and solve, leading to the verified form:

\[\frac{dy}{dx} = \frac{2}{1 - 3y^2}\]

This confirmation gives you the specific slope at any point on the curve, essential for understanding its overall shape and behavior.

**Steps for Verification:**

• Differentiate each term with respect to x.
• Use the chain rule for y terms.
• Rearrange to isolate \(\frac{dy}{dx}\).

To determine the slope of a curve at specific points, you can apply the derivative expression you've found and plug in the x and y values for those points. Here, we are asked to find the slope at each point where x equals 0, which corresponds to our earlier-found y-intercepts: (0, 0), (0, 1), and (0, -1).

Plug each of these points into the equation for the derivative \(\frac{dy}{dx} = \frac{2}{1 - 3y^2}\). For the point (0, 0), substituting y = 0 gives a slope of 2. For points (0, 1) and (0, -1), using y = 1 or y = -1 results in a slope of -1.

These calculated slopes give us a clear image of how the curve behaves at these points. A slope of 2 at (0, 0) suggests a steeper incline, while a slope of -1 at (0, 1) and (0, -1) indicates the curve is descending or ascending at these points, respectively. This is vital for visualizing the curve's local behavior at given points.

**Finding Slope Steps:**

• Identify y-intercepts to find specific points.
• Plug each point into the derivative.
• Calculate to find the slope for each point.