Problems 81– 88 require the following discussion of a secant line. The slope of the secant line containing the two points $(x,f(x\left)\right)$and $(x+h,f(x+h\left)\right)$on the graph of a function $y=f\left(x\right)$may be given as $m_{sec}=\frac{f(x+h)-f\left(x\right)}{h}$, where $h\ne 0$.

(a) Express the slope of the secant line of each function in terms of $x$and $h$. Be sure to simplify your answer.

(b) Find $m_{sec}$ for $h=0.5,0.1,$and $0.01$at $x=1$. What value does $m_{sec}$ approach as $h$approaches $0$?

(c) Find the equation for the secant line at$x=1$with $h=0.01$.

(d) Use a graphing utility to graph $f$and the secant line found in part (c) on the same viewing window.

$f\left(x\right)=-3x+2$

Consider the function.

$f\left(x\right)=-3x+2$

The slope of secant line of the given function is as follows:

$\begin{array}{rcl}{m}_{sec}& =& \frac{f(x+h)-f\left(x\right)}{h}\\ & =& \frac{-3(x+h)+2-(-3x+2)}{h}\\ & =& \frac{-3x-3h+2+3x-2}{h}\\ & =& \frac{-3h}{h}\\ & =& -3\end{array}$

From part (a), the value is $m_{sec}=-3$for all values of $h$since $m_{sec}$is constant in $h$.

Therefore, the value of $m_{sec}$for $h=0.5,0.1,$and $0.01$at $x=1$are $-3,-3,$and $-3$.

Thus, conclude that as $h$approaches to $0$, $m_{sec}$approaches $-3$.

From part (b), $m_{sec}=-3$at $x=1$and $h=0.01$.

The equation of a straight line with point $(x_1,y_1)$and the slope $m$is given by $y-y_1=m(x-x_1)$.

$\begin{array}{rcl}(1,f(1\left)\right)& =& (1,-3(1)+2)\\ & =& (1,-3+2)\\ & =& (1,-1)\end{array}$

Thus, the equation of a line with point $(1,f(1\left)\right)$and the slope $m=-3$is as follows:

$\begin{array}{rcl}y-(-1)& =& -3(x-1)\\ y+1& =& -3x+3\\ y& =& -3x+3-1\\ y& =& -3x+2\end{array}$

The function is $f\left(x\right)=-3x+2$and the secant line is $y=-3x+2$.

The graph of $f$and the secant line is as follows: