Chapter 2.4, Problem 12E

(a)

To determine

To find: The slope of the curve $y=x^2-3x-1$ at $x=0$.

Answer to Problem 12E

The slope of the curve $y=x^2-3x-1$ at $x=0$ is $-3$.

Explanation of Solution

Given information:

The curve is $y=x^2-3x-1$ and the point is $x=0$.

Calculation:

The formula for the slope of the curve $f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ is equal to $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$.

The slope of the curve $y=x^2-3x-1$ at $x=0$ is:

  $\begin{array}{c}\text{Slope}=\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h}\end{array}$

Substitute $h$ for $x$ in the function.

  $\begin{array}{l}f\left(x\right)=x^2-3x-1\\ f\left(h\right)=h^2-3h-1\end{array}$

Substitute $0$ for $x$ in the function.

  $\begin{array}{l}f\left(x\right)=x^2-3x-1\\ f\left(0\right)=0^2-3\times 0-1\\ f\left(0\right)=0-0-1\\ f\left(0\right)=-1\end{array}$

Substitute $h^2-3h-1$ for $f\left(h\right)$ and $-1$ for $f\left(0\right)$ in the formula for slope.

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

Therefore, the slope of the curve $y=x^2-3x-1$ at $x=0$ is $-3$.

(b)

To determine

To find: The equation of the tangent line to the curve $y=x^2-3x-1$ at $x=0$.

Answer to Problem 12E

The equation of the tangent line to the curve $y=x^2-3x-1$ at $x=0$ is $y=-\left(3x+1\right)$.

Explanation of Solution

Given information:

The curve is $y=x^2-3x-1$ and the point is $x=0$.

Calculation:

The equation of the tangent line to

  $y=f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ with slope at $x=a$ is:

  $y-f\left(a\right)=m\left(x-a\right)$

The function is $y=x^2-3x-1$ and as calculated in part (a), the value of $f\left(0\right)$ is equal to $-1$ . The value of slope of tangent to curve $y=x^2-3x-1$ at $x=0$ is $-3$.

Substitute $-3$ for $m$ , $0$ for $a$ and $-1$ for $f\left(0\right)$ in the equation of tangent.

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

Therefore, the equation of the tangent line to the curve $y=x^2-3x-1$ at $x=0$ is $y=-\left(3x+1\right)$.

(c)

To determine

To find: The equation to the normal line of the curve $y=x^2-3x-1$ at $x=0$.

Answer to Problem 12E

The equation of the normal line to the curve $y=x^2-3x-1$ at $x=0$ is $y=\frac{1}{3}x-1$.

Explanation of Solution

Given information:

The curve is $y=x^2-3x-1$ and the point is $x=0$.

Calculation:

The equation of the normal line of $y=f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ with slope at $x=a$ is:

  $y-f\left(a\right)=m\left(x-a\right)$

The function is $y=x^2-3x-1$ and as calculated in part (a), the value of $f\left(0\right)$ is $-1$ . The value of slope of tangent to curve $y=x^2-3x-1$ at $x=0$ is $-3$.

So, the slope of the normal to the curve $y=x^2-3x-1$ at $x=0$ is $\frac{1}{3}$.

Substitute $\frac{1}{3}$ for $m$ , $0$ for $a$ and $-1$ for $f\left(0\right)$ in the equation of normal.

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

Therefore, the equation of the normal line to the curve $y=x^2-3x-1$ at $x=0$ is $y=\frac{1}{3}x-1$.

(d)

To determine

To graph: The curve $y=x^2-3x-1$ , the tangent and normal line to the curve at $x=0$ in the same square viewing window.

Explanation of Solution

Given information:

The curve is $y=x^2-3x-1$ and the point is $x=0$.

Graph:

As calculated in part (b), the equation of the tangent line to the curve $y=x^2-3x-1$ at $x=0$ is $y=-\left(3x+1\right)$ . As calculated in part (c), the equation of normal line to the curve $y=x^2-3x-1$ at $x=0$ is $y=\frac{1}{3}x-1$.

To graph the curve, tangent and normal to the curve, follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $y=x^2-3x-1$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{-}\boxed{3}\boxed{\text{X}}\boxed{-}\boxed{1}$ . Now press the $\boxed{\text{ENTER}}$ button and enter right hand side of the equation $y=-\left(3x+1\right)$ after the symbol ${\text{Y}}_2$ . Enter the keystrokes $\boxed{(-)}\boxed{(}\boxed{3}\boxed{\text{X}}\boxed{+}\boxed{1}\boxed{)}$ . Now press the $\boxed{\text{ENTER}}$ button and enter right hand side of the equation $y=\frac{1}{3}x-1$ after the symbol ${\text{Y}}_3$ . Enter the keystrokes $\boxed{(}\boxed{(}\boxed{1}\boxed{÷}\boxed{3}\boxed{)}\boxed{\text{X}}\boxed{)}\boxed{-}\boxed{1}$.

The display will show the equations,

  $\begin{array}{l}{\text{Y}}_{\text{1}}=\text{X}^2-3\text{X}-1\\ {\text{Y}}_2=-\left(3\text{X}+1\right)\\ {\text{Y}}_3=\left(\left(1/3\right)\text{X}\right)-1\end{array}$

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown below.

  

Figure (1)

Interpretation: From the graph it can be seen that the curve, the tangent and the normal line intersect at the point $\left(0,-1\right)$.