91Ó°ÊÓ

The x-intercept of a curve represents the point where the curve crosses the x-axis. This intersection happens when the y-coordinate is zero. For parametric equations like ours, we evaluate these intersections using the given equations:

• The x-intercept occurs when \( y = 0 \).

In the given problem, the y-equation is \( y = t + 1 \). We set this to zero to find:\[ t + 1 = 0 \]Solving gives us \( t = -1 \). Now, substitute \( t = -1 \) back into the x-equation, \( x = t^2 - 2t \):\[ x = (-1)^2 - 2(-1) = 1 + 2 = 3 \]Thus, the x-intercept is at the point (3, 0). This tells us that the curve crosses the x-axis at the coordinate (3, 0). Therefore, when examining the behavior of this curve, knowing this point helps you visualize how and where the curve crosses the x-axis.

The y-intercept is the point where the curve crosses the y-axis. This happens when the x-coordinate is zero. In parametric equations, we determine this by setting the x-equation equal to zero.

• The y-intercept occurs when \( x = 0 \).

For our problem, we solve the x-equation \( x = t^2 - 2t \) by setting it to zero:\[ x = t^2 - 2t = 0 \]Factoring gives us:\[ t(t - 2) = 0 \]This results in two values for \( t \):

• \( t = 0 \)
• \( t = 2 \)

Substituting these into the y-equation \( y = t + 1 \):- For \( t = 0 \), \( y = 0 + 1 = 1 \).- For \( t = 2 \), \( y = 2 + 1 = 3 \).Hence, the y-intercepts are at points (0, 1) and (0, 3). These points describe where the curve intersects the y-axis, which helps in understanding the initial and end behavior of the curve on the y-axis within its parameter interval.

In parametric equations, the trajectory of the curve is defined for a specific range or interval of parameter values, known as the parameter interval. In our problem, the parameter interval is defined as:

• \(-2 \leq t < 4 \)

This interval gives the valid range for \( t \) that determines how the curve is traced on the plane. Let's break down why this interval is important:

Ensuring Values are Within Bounds

Specific values of \( t \) are necessary to determine the x- and y-intercepts:- We calculated \( t = 0 \), \( t = 2 \), and \( t = -1 \) for the intercept points.- We must ensure these values fall within our defined interval.

Implications of the Interval

- This interval is crucial in understanding the entire behavior of the curve.- Since \( t \) can have values starting from \(-2\) up to but not including \(4\), the curve will only trace its path for this range of \( t \).Understanding the parameter interval is essential when analyzing parametric equations. It enhances comprehension of how the curve behaves and changes as the parameter \( t \) varies within this specific range.