Chapter 1.4, Problem 33E

To determine

To find: The parameterization for the part of the graph lies in Quadrant I.

Answer to Problem 33E

A possible parameterization is $x=t$ and $y=t^2+2t+2$ for $t>0$.

Explanation of Solution

Given information: The given equation is $y=x^2+2x+2$.

Calculation:

Use the following step to graph the parametric equations by graphing calculator.

Step 1: Press “ON” button to open the calculator. Press $\boxed{\text{MODE}}$ button, and choose FUNC option and press $\boxed{\text{ENTER}}$.

Step 2: Press $\boxed{\text{Y=}}$ and enter the right hand side of the equation equal to ${\text{Y}}_1$ by the keystrokes $\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{+}\boxed{2}\boxed{\text{X}}\boxed{+}\boxed{2}$.

Step 3: Press $\boxed{\text{GRAPH}}$ to obtain the graph of the as shown below.

  

As sown in the above graph, the graph of $y=x^2+2x+2$ lies in the Quadrant I for all $x>0$.

To find the parameterization substitute $t$ for $x$ in the equation $y=x^2+2x+2$.

  $y=t^2+2t+2$

Therefore, a possible parameterization is $x=t$ and $y=t^2+2t+2$ for $t>0$.