Chapter 0.4, Problem 31E

To determine

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

Answer to Problem 31E

  $-5\le t<-3$

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 III does not include the axes.

So, the portion of graph in the third quadrant consists of line segment between the points $\left(-2,-6\right)$ (this point is inclusive) and $\left(0,-4\right)$ (this point is not inclusive as it lies of axis)

Now, first consider the point $\left(-2,-6\right)$ .

Substitute $y=-6$ into the given equation $y=t-1$ to find value of t .

  $\begin{array}{c}y=t-1\\ -6=t-1\\ t=-6+1\\ t=-5\end{array}$

Now, consider the point $\left(0,-4\right)$ .

Substitute $y=-4$ into the given equation $y=t-1$ to find value of t .

  $\begin{array}{c}y=t-1\\ -4=t-1\\ t=-4+1\\ t=-3\end{array}$

Therefore, the required values of t are $-5\le t<-3$ .