Chapter ISG, Problem 3.3.7P

To determine

To Prove: $y=-ax+b$ grows by an equal difference over any interval $k$ .

Explanation of Solution

Given:

  $y=-ax+b$ where $a\ne 0$ , $k$ where $k\ne 0$

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

If $x=x_0$ , then $y=-a{x}_0+b$

If $x=x_0+k$ , then $y=-a(x_0+k)+b\Rightarrow y=-a{x}_0-ak+b$

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

The difference between the $y$ values is $-a{x}_0-ak+b-(-a{x}_0+b)=-ak$ .

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

Therefore, completing the table:

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

So, the linear function $y=-ax+b$ grows by an equal difference over any interval $k$ because the difference at each interval of $k$ is equal to $-ak$ .