Chapter 2.4, Problem 9E

(a)

To determine

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

Answer to Problem 9E

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

Explanation of Solution

Given information:

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

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 $y=x^2$ at $x=-2$ is:

  $\text{Slope}=\underset{h\to 0}{\lim }\frac{f\left(-2+h\right)-f\left(-2\right)}{h}$

Substitute $-2+h$ for $x$ in the function.

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

Substitute $-2$ for $x$ in the function.

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

Substitute $4+h^2-4h$ for $f\left(-2+h\right)$ and $4$ for $f\left(-2\right)$ in the formula for slope.

  $\begin{array}{c}\text{Slope}=\underset{h\to 0}{\lim }\frac{4+h^2-4h-4}{h}\\ =\underset{h\to 0}{\lim }\frac{h\left(h-4\right)}{h}\\ =\underset{h\to 0}{\lim }\left(h-4\right)\\ =-4\end{array}$

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

(b)

To determine

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

Answer to Problem 9E

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

Explanation of Solution

Given information:

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

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:

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

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

Substitute $-4$ for $m$ , $-2$ for $a$ and $4$ for $f\left(-2\right)$ in the equation of tangent.

  $\begin{array}{c}y-f\left(-2\right)=-4\left(x-\left(-2\right)\right)\\ y-4=-4\left(x+2\right)\\ y-4=-4x-8\\ y=-4\left(x+1\right)\end{array}$

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

(c)

To determine

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

Answer to Problem 9E

The equation of the normal line to the curve $y=x^2$ at $x=-2$ is $y=\frac{1}{4}x+\frac{9}{2}$.

Explanation of Solution

Given information:

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

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 $f\left(-2\right)$ is equal to $4$ . The value of slope of tangent to curve $y=x^2$ at $x=-2$ is $-4$.

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

Substitute $\frac{1}{4}$ for $m$ , $-2$ for $a$ and $4$ for $f\left(-2\right)$ in the equation of normal.

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

Further simplify.

  $\begin{array}{c}y=\frac{1}{4}x+\frac{1}{2}+4\\ y=\frac{1}{4}x+\frac{9}{2}\end{array}$

Therefore, the equation of the normal line to the curve $y=x^2$ at $x=-2$ is $y=\frac{1}{4}x+\frac{9}{2}$.

(d)

To determine

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

Explanation of Solution

Given information:

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

Graph:

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

As calculated in part (c), the equation of normal line to the curve $y=x^2$ at $x=-2$ is equal to $y=\frac{1}{4}x+\frac{9}{2}$.

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 $y=x^2$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{\text{X}}\boxed{^}\boxed{2}$ . Now, press the $\boxed{\text{ENTER}}$ button and enter right hand side of the equation $y=-4\left(x+1\right)$ after the symbol ${\text{Y}}_2$ . Enter the keystrokes $\boxed{(-)}\boxed{4}\boxed{(}\boxed{\text{X}}\boxed{+}\boxed{1}\boxed{)}$ . Now, press the $\boxed{\text{ENTER}}$ button and enter right hand side of the equation $y=\frac{1}{4}x+\frac{9}{2}$ after the symbol ${\text{Y}}_3$ . Enter the keystrokes $\boxed{(}\boxed{\text{X}}\boxed{÷}\boxed{4}\boxed{)}\boxed{+}\boxed{(}\boxed{9}\boxed{÷}\boxed{2}\boxed{)}$.

The display will show the equations,

  $\begin{array}{l}{\text{Y}}_{\text{1}}=\text{X}^2\\ {\text{Y}}_2=-4\left(\text{X}+1\right)\\ {\text{Y}}_3=\left(\frac{\text{X}}{4}\right)+\left(\frac{9}{2}\right)\end{array}$

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 $\left(-2,4\right)$.