Chapter 12.3, Problem 28E

To determine

Sketch the graph of the given function and the tangent line at point $\left(1,f(1)\right)$ and find the slope of the tangent line.

Answer to Problem 28E

The approximate slope of the curve at point $(1,3)$ is $\text{Slope}=3$

Explanation of Solution

Given:

The function given is:

The given function is:

  $f(x)=x^2-2x+1$

Since, the tangent is drawn to be on the point $\left(1,f(1)\right)$ .

  $\begin{array}{cc}f(1)& =1^2-2\times 1+1\\ & =2-2\\ & =0\end{array}$

The point $(1,3)$ as marked in the graph below. Now, draw the tangent at $\left(1,3\right)$

  

By the above graph the slope at point $(1,3)$ will be as,

  $\text{Slope}=\frac{\text{change in y}}{\text{change in x}}$

Now, from the above graph, the tangent is at $x=1$ when $y=3$ and $x=0$ when $y=0$ . Hence by the above formula of slope of the tangent

  $\begin{array}{l}\text{Slope}=\frac{3}{1}=3\\ \end{array}$

Therefore, the approximate slope of the curve at point $(1,3)$ is $\text{Slope}=3$