Q 85. - Precalculus Enhanced with Graphing Utilities

Question

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 msec for $h = 0.5, 0.1,$ and $0.01$ at $x = 1$. What value does msec 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.

$f\left(x\right)=2x^2-3x+1$

Step-by-Step Solution

Answer

Part (a) The slope of the secant line of the given function is $m_{sec}=2h+4x-3$.

Part (b) The value of $m_{sec}$ for $h=0.5,0.1,$ and $0.01$ are $2,1.2,$ and $1.02$. The value of $m_{sec}$ is $1$ when $h$ approahes $0$.

Part (c) The equation for the secant line at $x=1$ with $h=0.01$ is $\begin{array}{rcl}y& =& 1.02x-1.02\end{array}$.

Part (d) The graph of the given function and the secant line is shown below:

Part (a) Step 1. Given information

Consider the function.

$f\left(x\right)=2x^2-3x+1$

Part (a) Step 2. Express the slope of the secant line of the given function in terms of x and h .

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

$\begin{array}{rcl}{m}_{sec}& =& \frac{f(x+h)-f\left(x\right)}{h}\\ & =& \frac{2{(x+h)}^2-3(x+h)+1-(2x^2-3x+1)}{h}\\ & =& \frac{2x^2+2h^2+4xh-3x-3h+1-2x^2+3x-1}{h}\\ & =& \frac{2h^2+4xh-3h}{h}\\ & =& 2h+4x-3\end{array}$

Part (b) Step 1. Determine the of m s e c for h = 0 . 5 , 0 . 1 , and 0 . 01 at x = 1 . Find the value of m s e c when h approaches 0 .

From part (a), the value is $m_{sec}=2h+4x-3$.

The value of $m_{sec}$ at $x=1$ with $h=0.5$ is as follows:

 $\begin{array}{rcl}{m}_{sec}& =& 2(0.5)+4\left(1\right)-3\\ & =& 2\end{array}$

The value of $m_{sec}$ at $x=1$ with $h=0.1$ is as follows:

 $\begin{array}{rcl}{m}_{sec}& =& 2(0.1)+4\left(1\right)-3\\ & =& 1.2\end{array}$

The value of $m_{sec}$ at $x=1$ with $h=0.01$ is as follows:

$\begin{array}{rcl}{m}_{sec}& =& 2(0.01)+4\left(1\right)-3\\ & =& 1.02\end{array}$

Thus, conclude that as $h$ approaches to $0$, $m_{sec}$ approaches $1$.

Part (c) Step 1. Determine the equation for the secant line at x = 1 with h = 0 . 01 .

From part (b), $\begin{array}{rcl}{m}_{sec}& =& 1.02\end{array}$ 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 $y-y_1=m(x-x_1)$.

 $\begin{array}{rcl}(1,f(1\left)\right)& =& (1,{2\left(1\right)}^2-3(1)+1)\\ & =& (1,2-3+1)\\ & =& (1,0)\end{array}$

Thus, the equation of a line with point $(1,f(1\left)\right)$ and the slope $m=1.02$ is as follows:

$\begin{array}{rcl}y-0& =& 1.02(x-1)\\ y& =& 1.02x-1.02\end{array}$

Part(d) Step 1. Graph f and the secant line found in part (c) by applying a graphing utility.

The function is $f\left(x\right)=2x^2-3x+1$ and the secant line is $\begin{array}{rcl}y& =& 1.02x-1.02\end{array}$.

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