Question

In Problems 9–20, find the slope of the tangent line to the graph of f at the given point. Graphf and the tangent line.

$f\left(x\right)=x^3-x^2$at$\left(1,0\right)$

Short answer

The slope of tangent is $1$.

The graph is as follows,

Step-by-step solution

Step 1. Given Information 

We are given a function,

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

We have to find the slope of the tangent line and graph the function and tangent line.

Step 2. Finding the slope 

The slope is given by,

$m_{\tan }=\underset{x\to e}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$

Putting the values, we get

$$\begin{gathered} m=\underset{x\to 1}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-\left(1\right)} \\ =\underset{x\to 1}{\lim }\frac{x^3-x^2-0}{x-1} \\ =\underset{x\to 1}{\lim }\frac{x^2(x-1)}{x-1} \\ =\underset{x\to 1}{\lim }{x}^2 \\ =1^2 \\ =1 \end{gathered}$$

Hence the slope of tangent is $1$.

Step 3. Graphing the function 

The equation of tangent is given by,

$$\begin{gathered} y-y_1=m\left(x-x_1\right) \\ y-0=1(x-1) \\ y=x-1 \end{gathered}$$

The graph is as follows,