Problem 29 - Precalculus with Calculus Previews

Question

Find the \(x\) - and \(y\) -intercepts of the given curves. $$ x=t^{2}-2 t, y=t+1,-2 \leq t<4 $$

Step-by-Step Solution

Answer

Y-intercepts: (0,1) and (0,3); X-intercept: (3,0).

Step 1: Understand the problem

We are given a set of parametric equations: \( x = t^2 - 2t \) and \( y = t + 1 \) with the parameter \( t \) defined on the interval \(-2 \leq t < 4\). We need to find the \(x\)- and \(y\)-intercepts of these equations.

Step 2: Find the y-intercept

The \(y\)-intercept occurs when \(x = 0\). So, first set \(x = t^2 - 2t = 0\). Factor to get \(t(t - 2) = 0\), giving solutions \(t = 0\) and \(t = 2\). Calculate \(y\) for these \(t\) values. When \(t = 0\), \(y = 0 + 1 = 1\). When \(t = 2\), \(y = 2 + 1 = 3\). Thus, the \(y\)-intercepts are \((0, 1)\) and \((0, 3)\).

Step 3: Find the x-intercept

The \(x\)-intercept occurs when \(y = 0\). Set \(y = t + 1 = 0\), solving \(t = -1\). Substitute \(t = -1\) into \(x = t^2 - 2t\) to get \(x = (-1)^2 - 2(-1) = 1 + 2 = 3\). Thus, the \(x\)-intercept is \((3, 0)\).

Step 4: Confirm intercepts within bounds

We must confirm that the values \(t = 0\), \(t = 2\), and \(t = -1\) fall within the given interval \(-2 \leq t < 4\): \(t = 0\) and \(t = 2\) are within bounds. \(t = -1\) is also valid. Thus, all calculated intercepts are appropriate.

Step 5: Conclusion

The \(y\)-intercepts of the parametric curve are \((0, 1)\) and \((0, 3)\). The \(x\)-intercept is \((3, 0)\). These points represent where the curve crosses the axes within the defined interval of \(t\).

Key Concepts

Understanding the x-intercept

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.

Identifying the y-intercept

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.

Exploring the 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.