Question

and identify the graph of the parametric curve. in exercises 11–26, use an algebraic method to eliminate the parameter 12. ( x = 2 - 3t, y = 5 + t )

Explanation:

Step1: Solve for \( t \) from the \( y \)-equation

From \( y = 5 + t \), we can isolate \( t \) by subtracting 5 from both sides. So, \( t = y - 5 \).

Step2: Substitute \( t \) into the \( x \)-equation

We have \( x = 2 - 3t \). Substitute \( t = y - 5 \) into this equation:
\( x = 2 - 3(y - 5) \)

Step3: Simplify the equation

First, distribute the -3: \( x = 2 - 3y + 15 \)
Then, combine like terms: \( x = 17 - 3y \)
We can rearrange this into slope - intercept form (\( y = mx + b \)) or standard form. Let's rearrange it to solve for \( y \):
Add \( 3y \) to both sides: \( 3y+x = 17 \)
Subtract \( x \) from both sides: \( 3y=-x + 17 \)
Divide both sides by 3: \( y=-\frac{1}{3}x+\frac{17}{3} \)

Answer:

The rectangular equation after eliminating the parameter is \( y = -\frac{1}{3}x+\frac{17}{3} \) (or equivalent forms like \( x + 3y=17 \)), and the graph of the parametric curve is a straight line with a slope of \( -\frac{1}{3} \) and a \( y \)-intercept of \( \frac{17}{3} \).