Chapter 1.4, Problem 12E

a.

To determine

To find: The slope of the curve at the given point.

Answer to Problem 12E

  $-3$ .

Explanation of Solution

Given information:

The function is $y=x^2-3x-1$ at $x=0$ .

Concept used:

Slope of any function f at any point a is given by: $f^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

Calculation:

Slope $y=x^2-3x-1$ at $x=0$ will be:

  $\begin{array}{c}{f}^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{h^2-3h-1-0^2+0^2+1}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{h^2-3h}{h}\right)\\ =\underset{h\to 0}{\lim }\left(h-3\right)\\ =-3\end{array}$

Conclusion:

The slope of the curve at the given point is $-3$ .

b.

To determine

To find: the equation of the tangent line.

Answer to Problem 12E

  $y=-3x-1$ .

Explanation of Solution

Given information:

The function is $y=x^2-3x-1$ at $x=0$ .

Concept used:

For the equation for the tangent line to $f\left(x\right)$ at $x=x_0$ , first find the value of y at $x=x_0$ .

Then find the slope of the tangent line at $x=x_0$ by finding the derivative of the function, evaluated at $x=x_0$ .

The equation of the tangent line is $y-y_0=m\left(x-x_0\right)$ .

Calculation:

From part (a) it is obtained that slope at $x=0$ is $-3$ .

Find $f\left(0\right)$ for $y_0$ .

  $\begin{array}{c}{y}_0=0^2-0-1\\ =-1\end{array}$

Hence $\left(x_0,y_0\right)=\left(0,-1\right)$ .

Put $\left(x_0,y_0\right)=\left(0,-1\right)$ and $m=-3$ in the equation of the tangent line, that is, $y-y_0=m\left(x-x_0\right)$ .

  $\begin{array}{c}y-\left(-1\right)=-3\left(x-\left(0\right)\right)\\ y+1=-3\left(x\right)\\ y=-3x-1\end{array}$

Hence the equation of tangent line is $y=-3x-1$ .

c.

To determine

To find: The equation of the normal.

Answer to Problem 12E

  $y=\frac{1}{3}x-1$ .

Explanation of Solution

Given information:

The function is $y=x^2-3x-1$ at $x=0$ .

Concept used:

If the slope of a line is m, then the slope of the normal will be $-\frac{1}{m}$ . Also, the normal passes through the point $\left(x_0,y_0\right)$ . So, the equation of the normal will be:

  $y-y_0=-\frac{1}{m}\left(x-x_0\right)$ .

Calculation:

For $\left(x_0,y_0\right)=\left(0,-1\right)$ and $m=-3$ , the equation of the normal is:

  $\begin{array}{c}y-y_0=-\frac{1}{m}\left(x-x_0\right)\\ y-\left(-1\right)=-\left(-\frac{1}{3}\right)\left(x-0\right)\\ y=\frac{1}{3}x-1\end{array}$

Hence the equation of the normal is $y=\frac{1}{3}x-1$ .

d.

To determine

To graph: The curve, tangent line and the normal in the same square viewing window.

Answer to Problem 12E

  

Explanation of Solution

Given information:

The function is $y=x^2-3x-1$ at $x=0$ .

Graph:

Using graphing calculator:

  

Interpretation:

From the graph it is clear that all the curve passes through the point $\left(0,-1\right)$ .