Chapter 3, Problem 1P

(a)

To determine

To find: The quadratic polynomial that best fits the given data.

Answer to Problem 1P

The quadratic polynomial that best fits the given data is equal to $y=-0.2754x^2+19.7485x-273.5523$ .

Explanation of Solution

Given information:

Car tires need to be inflated properly. Over inflation or under inflation can cause premature tread ware. The data and scatter plot given below show the tire life for different inflation values for a certain type of tire.

Pressure $\left(\text{lb}/{\text{in}}^2\right)$	Tire life $\left(\text{mi}\right)$
$26$	$50000$
$28$	$66000$
$31$	$78000$
$35$	$81000$
$38$	$74000$
$42$	$70000$
$45$	$59000$

  

Calculation:

Represent the tire life per thousand.

For the quadratic model of data follow the steps below:

First press the $\boxed{2\text{nd}}$ key and then $\boxed{\text{Y}=}$ and make sure that Plot $1$ is ON and choose the scatter plot icon in type.

Go to ${\text{Y}}_1$ and then press the key $\boxed{\text{STAT}}$ and then in edit enter the data in ${\text{L}}_1$ and ${\text{L}}_2$ .

Now enter the keystrokes $\boxed{\text{STAT}}\boxed{\to }\boxed{5}$ and press the $\boxed{\text{ENTER}}$ key for the regression.

  

Figure(1)

Therefore, the quadratic polynomial that best fits the given data is equal to $y=-0.2754x^2+19.7485x-273.5523$ .

(b)

To determine

To graph: The scatter plot of the given data and the quadratic polynomial obtained from part(a) in the same viewing window.

Explanation of Solution

Given information:

Car tires need to be inflated properly. Over inflation or under inflation can cause premature tread ware. The data and scatter plot given below show the tire life for different inflation values for a certain type of tire.

Pressure $\left(\text{lb}/{\text{in}}^2\right)$	Tire life $\left(\text{mi}\right)$
$26$	$50000$
$28$	$66000$
$31$	$78000$
$35$	$81000$
$38$	$74000$
$42$	$70000$
$45$	$59000$

  

Graph:

To graph the points on scatter plot, follow the steps using graphing calculator.

First press the $\boxed{2\text{nd}}$ key and then $\boxed{\text{Y}=}$ and make sure that Plot $1$ is ON and choose the scatter plot icon in type.

Go to ${\text{Y}}_1$ and then press the key $\boxed{\text{STAT}}$ and then in edit enter the data in ${\text{L}}_1$ and ${\text{L}}_2$ .

Now press the key $\boxed{\text{Y}=}$ and go to ${\text{Y}}_2$ and enter the right hand side of the function $y=-0.2754x^2+19.7485x-273.5523$ . Enter the keystrokes $\boxed{(-)}\boxed{0.2754}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{+}\boxed{19.7485}\boxed{\text{X}}\boxed{-}\boxed{273.5523}$ .

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph in standard window as shown in Figure (2).

  

Figure (2)

(c)

To determine

To find: The pressure that gives the longest tire life with the help of results in part(b).

Answer to Problem 1P

The pressure that gives longest tire life is $35.85\text{lb}/{\text{in}}^2$ .

Explanation of Solution

Given information:

Car tires need to be inflated properly. Over inflation or under inflation can cause premature tread ware. The data and scatter plot given below show the tire life for different inflation values for a certain type of tire.

Pressure $\left(\text{lb}/{\text{in}}^2\right)$	Tire life $\left(\text{mi}\right)$
$26$	$50000$
$28$	$66000$
$31$	$78000$
$35$	$81000$
$38$	$74000$
$42$	$70000$
$45$	$59000$

  

Calculation:

As observed from the graph of the function in part(b), the function achieves the maximum value at $x=35.85$ .

So, the maximum long life of tires is at the tire pressure of maximum value.

Therefore, the pressure that gives longest tire life is $35.85\text{lb}/{\text{in}}^2$ .