Chapter 0.4, Problem 29E

To determine

To find: the value of t that produces the graph in the first quadrant.

Answer to Problem 29E

  $1<t<3$

Explanation of Solution

Given information:

The equation:

  $x=3-\left|t\right|,y=t-1,-\text{5}\le t\le 5$

The graph:

  

First it need to remember that Quadrant I does not include the axes.

So, the portion of graph in the first quadrant consists of line segment between the points $\left(2,0\right)$ and $\left(0,2\right)$ (Both points not inclusive as they lie of the axes.)

Now, first consider the point $\left(2,0\right)$ .

Substitute $y=0$ into the given equation $y=t-1$ to find value of t .

  $\begin{array}{c}y=t-1\\ 0=t-1\\ t=0+1\\ t=1\end{array}$

Now, consider the point $\left(0,2\right)$ .

Substitute $y=2$ into the given equation $y=t-1$ to find value of t .

  $\begin{array}{c}y=t-1\\ 2=t-1\\ t=2+1\\ t=3\end{array}$

Therefore, the required values of t are $1<t<3$ .