Chapter 2.4, Problem 10E

(a)

To determine

To find: The slope of the curve $y=x^2-4x$ at $x=1$.

Answer to Problem 10E

The slope of the curve $y=x^2-4x$ at $x=1$ is $-2$.

Explanation of Solution

Given information:

The curve is $y=x^2-4x$ and the point is $x=1$.

Calculation:

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

The slope of the curve $y=x^2-4x$ at $x=1$ is:

  $\text{Slope}=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}$

Substitute $1+h$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=x^2-4x\\ f\left(1+h\right)={\left(1+h\right)}^2-4\left(1+h\right)\\ f\left(1+h\right)=1+h^2+2h-4-4h\\ f\left(1+h\right)=h^2-2h-3\end{array}$

Substitute $1$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=x^2-4x\\ f\left(1\right)={\left(1\right)}^2-4\left(1\right)\\ f\left(1\right)=1-4\\ f\left(1\right)=-3\end{array}$

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

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

Therefore, the slope of the curve $y=x^2-4x$ at $x=1$ is $-2$.

(b)

To determine

To find: The equation of the tangent line to the curve $y=x^2-4x$ at $x=1$.

Answer to Problem 10E

The equation of the tangent line to the curve $y=x^2-4x$ at $x=1$ is $y=-\left(2x+1\right)$.

Explanation of Solution

Given information:

The curve is $y=x^2-4x$ and the point is $x=1$.

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 $y=x^2$ and as calculated in part (a), the value of $f\left(1\right)$ is equal to $-3$ . Also, the value of slope of tangent to curve $y=x^2-4x$ at $x=1$ is $-2$.

Substitute $-2$ for $m$ , $1$ for $a$ and $-3$ for $f\left(1\right)$ in the equation of tangent.

  $\begin{array}{c}y-f\left(1\right)=-2\left(x-1\right)\\ y-\left(-3\right)=-2x+2\\ y+3=-2x+2\\ y=-\left(2x+1\right)\end{array}$

Therefore, the equation of the tangent line to the curve $y=x^2-4x$ at $x=1$ is $y=-\left(2x+1\right)$.

(c)

To determine

To find: The equation to the normal line of the curve $y=x^2-4x$ at $x=1$.

Answer to Problem 10E

The equation of the normal line to the curve $y=x^2-4x$ at $x=1$ is $y=\frac{1}{2}\left(x-7\right)$.

Explanation of Solution

Given information:

The curve is $y=x^2-4x$ and the point is $x=1$.

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:

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

As calculated in part (a), the value of $f\left(1\right)$ is equal to $-3$ . The value of slope of tangent to curve $y=x^2-4x$ at $x=1$ is equal to $-2$.

So, the slope of the normal to the curve $y=x^2-4x$ at $x=1$ is $\frac{1}{2}$.

Substitute $\frac{1}{2}$ for $m$ , $1$ for $a$ and $-3$ for $f\left(1\right)$ in the equation of tangent.

  $\begin{array}{c}y-f\left(1\right)=\frac{1}{2}\left(x-1\right)\\ y-\left(-3\right)=\frac{1}{2}\left(x-1\right)\\ y+3=\frac{1}{2}x-\frac{1}{2}\end{array}$

Further simplify.

  $\begin{array}{c}y=\frac{1}{2}x-\frac{1}{2}-3\\ y=\frac{1}{2}\left(x-7\right)\end{array}$

Therefore, the equation of the normal line to the curve $y=x^2-4x$ at $x=1$ is $y=\frac{1}{2}\left(x-7\right)$.

(d)

To determine

To graph: The curve $y=x^2-4x$ , the tangent and normal line to the curve at $x=1$ in the same square viewing window.

Explanation of Solution

Given information:

The curve is $y=x^2-4x$ and the point is $x=1$.

Graph:

As calculated in part (b), the equation of the tangent line to the curve $y=x^2-4x$ at $x=1$ is equal to $y=-\left(2x+1\right)$.

As calculated in part (c), the equation of normal line to the curve $y=x^2-4x$ at $x=1$ is equal to $y=\frac{1}{2}\left(x-7\right)$.

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 $y=x^2-4x$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{-}\boxed{4}\boxed{\text{X}}$ . Now, press the $\boxed{\text{ENTER}}$ button and enter right hand side of the equation $y=-\left(2x+1\right)$ after the symbol ${\text{Y}}_2$ . Enter the keystrokes $\boxed{(-)}\boxed{(}\boxed{2}\boxed{\text{X}}\boxed{+}\boxed{1}\boxed{)}$ . Now, press the $\boxed{\text{ENTER}}$ button and enter right hand side of the equation $y=\frac{1}{2}\left(x-7\right)$ after the symbol ${\text{Y}}_3$ . Enter the keystrokes $\boxed{(}\boxed{1}\boxed{÷}\boxed{2}\boxed{)}\boxed{(}\boxed{\text{X}}\boxed{-}\boxed{7}\boxed{)}$.

The display will show the equations,

  $\begin{array}{l}{\text{Y}}_{\text{1}}=\text{X}^2-4\text{X}\\ {\text{Y}}_2=-\left(\text{2X}+1\right)\\ {\text{Y}}_3=\left(1/2\right)\left(\text{X}-7\right)\end{array}$

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 $\left(1,-3\right)$.