Chapter ISG, Problem 3.3.6P

To determine

To find: The complete table and how does the value of $y$ change over each interval $k$ .

Answer to Problem 3.3.6P

  $\begin{array}{cccc}x& y=-ax+b& y& Change\\ x_0& -a(x_0)+b& -a{x}_0+b& ----\\ x_0+k& -a(x_0+k)+b& -a{x}_0-ak+b& -ak\\ x_0+2k& -a(x_0+2k)+b& -a{x}_0-2ak+b& -ak\\ x_0+3k& -a(x_0+3k)+b& -a{x}_0-3ak+b& -ak\\ x_0+4k& -a(x_0+4k)+b& -a{x}_0-4ak+b& -ak\end{array}$

Explanation of Solution

Given:

  

  $y=-ax+b$

  $\begin{array}{cccc}x& y=-ax+b& y& Change\\ x_0& -a(x_0)+b& -a{x}_0+b& ----\\ x_0+k& -a(x_0+k)+b& -a{x}_0-ak+b& -ak\\ x_0+2k& & & \\ x_0+3k& & & \\ x_0+4k& & & \end{array}$

To find the change in $y$ over each interval $k$ , $x$ is replaced with value $x_0+k$ in the function $y=-ax+b$

Where $x_0$ is the initial value of $x$ .

It can be observed from the given values that, $y$ is decreasing with the value $-ak$ with an increase of $k$ in $x$ .

Therefore, the following table can be made:

  $\begin{array}{cccc}x& y=-ax+b& y& Change\\ x_0& -a(x_0)+b& -a{x}_0+b& ----\\ x_0+k& -a(x_0+k)+b& -a{x}_0-ak+b& -ak\\ x_0+2k& -a(x_0+2k)+b& -a{x}_0-2ak+b& -ak\\ x_0+3k& -a(x_0+3k)+b& -a{x}_0-3ak+b& -ak\\ x_0+4k& -a(x_0+4k)+b& -a{x}_0-4ak+b& -ak\end{array}$

The labeled $y$

values are as follows: