Question

consider the following parametric equations:

$x = \cos(\theta)$ and $y = 3\cos(\theta)+1$

step 2 of 2 : determine the domain and range of the equation obtained by eliminating the parameter.
please write your answer in interval notation.

Explanation:

Step1: Eliminate the parameter

Given \(x = \cos(\theta)\) and \(y=3\cos(\theta)+1\). Substitute \(x\) into the equation for \(y\). We get \(y = 3x+1\). But we know that \(- 1\leqslant\cos(\theta)\leqslant1\). Since \(x=\cos(\theta)\), the domain of \(x\) is based on the range of the cosine function.

Step2: Find the domain

Since \(x = \cos(\theta)\) and \(-1\leqslant\cos(\theta)\leqslant1\), the domain of \(x\) (for the function \(y = 3x + 1\) with the parameter - based constraint) is \([-1,1]\).

Step3: Find the range

We use the function \(y=3x + 1\) and the domain \(x\in[-1,1]\).
When \(x=-1\), \(y=3\times(-1)+1=-3 + 1=-2\).
When \(x = 1\), \(y=3\times1+1=3 + 1=4\).
Since \(y = 3x+1\) is a linear function (\(y=mx + b\) with \(m = 3>0\), it is increasing), the range of \(y\) is found by evaluating the function at the endpoints of the domain.

Answer:

Domain: \([-1,1]\), Range: \([-2,4]\)