Chapter ISG, Problem 3.3.8P

To determine

To Discuss: Relation between differences in the $y$ values over equal intervals to the slope and rate of change of the function.

Answer to Problem 3.3.8P

Slope and the rate of change of the function is equal to the coefficient of $x$ in the linear equation of the type $y=-ax+b$ .

Explanation of Solution

Given:

Linear functions grow by equal differences over equal intervals.

Considering the linear function: $y=-ax+b$

Putting $x=x_0\Rightarrow$ $y=-a{x}_0+b$

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

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 $-ak$ .

Slope of a linear function is given as:

  $Slope=\frac{y_2-y_1}{x_2-x_1}\Rightarrow \frac{-ak}{k}=-a$

Rate of change of a function is given by its first derivative $\frac{dy}{dx}$ .

Differentiating the function $y=-ax+b$

  $\Rightarrow \frac{dy}{dx}=-a$ .

So, the slope and the rate of change of the function is equal to the coefficient of $x$ in the linear equation of the type $y=-ax+b$ .