Question

graph the curve whose parametric equations are given and show its orientation. find the rectangular equation curve

$x = 2t + 3$, $y = t + 5$; $0 \leq t \leq 4$

choose the correct graph below.

a.
b.
c.
d.

Explanation:

Step1: Eliminate the parameter \(t\)

From \(x = 2t+3\), we can solve for \(t\): \(t=\frac{x - 3}{2}\).
Substitute \(t=\frac{x - 3}{2}\) into \(y=t + 5\).
So \(y=\frac{x - 3}{2}+5\).
Simplify the equation:

$$
LATEXBLOCK0
$$

Step2: Find the range of \(x\) and \(y\) values

When \(t = 0\):
\(x=2(0)+3=3\), \(y=0 + 5=5\)
When \(t = 4\):
\(x=2(4)+3=11\), \(y=4 + 5=9\)

Answer:

The rectangular equation is \(x-2y=-7\) with \(3\leq x\leq11\) and \(5\leq y\leq9\). To determine the graph, we note the line segment starts at the point \((3,5)\) and ends at the point \((11,9)\). (Since the options are not fully described with coordinates, but based on the range of \(x\) (from \(3\) to \(11\)) and \(y\) (from \(5\) to \(9\)) and the slope \(m=\frac{1}{2}\) (from \(y=\frac{1}{2}x+\frac{7}{2}\)), the correct graph should be a line - segment within this \(x\) and \(y\) range).