Chapter 9.1, Problem 67PPS

(a)

To determine

The related function for each equation . Complete the first two columns of the table.

Explanation of Solution

Given Information:

The table is :

  

Calculation:

Writing the related function :

First make the equation with right hand side = 0 and then we get the related function.

Equation	Related Function
$x^2-x=12$	$y=x^2-x-12$
$x^2+8x=9$	$y=x^2+8x-9$
$x^2=14x-24$	$y=x^2-14x+24$
$x^2+16x=-28$	$y=x^2+16x+28$

(b)

To determine

To Graph:

Each related function with graphing calculator.

Explanation of Solution

Given Information:

From part (a), we have the table for each related function:

Equation	Related Function
$x^2-x=12$	$y=x^2-x-12$
$x^2+8x=9$	$y=x^2+8x-9$
$x^2=14x-24$	$y=x^2-14x+24$
$x^2+16x=-28$	$y=x^2+16x+28$

The related functions:

1. $y=x^2-x-12$
2. $y=x^2+8x-9$
3. $y=x^2-14x+24$
4. $y=x^2+16x+28$

Graph:

Using TI-83 Plus Graphing Calculator , we graph the given related functions:

1)

• Let Y1 = $x^2-x-12$

    

• Under the ZOOM option , select 6) and adjust the viewing option to standard:

  

• Use the WINDOW option to adjust the scaling of the graph :

  

• Use the GRAPH option to view the graph :

  

Obtain rest of the graphs of the related functions similarly:

2) Let Y1= $x^2+8x-9$ and adjust the scaling of the graph using WINDOW option , then obtain the graph using GRAPH option:

  

3) Let Y1= $x^2-14x+24$ and adjust the scaling of the graph using WINDOW option , then obtain the graph using GRAPH option:

  

4) Let Y1= $x^2+16x+28$ and adjust the scaling of the graph using WINDOW option , then obtain the graph using GRAPH option:

  

(c)

To determine

To Find:

Use the table feature on your calculator to determine the zeros of each related function. Record the zeroes in the table below . Also record the values of the function one unit less than and one unit more than each zero.

Explanation of Solution

Given Information:

The table is :

  

From part (a), the table is :

Equation	Related Function	Zeroes	y-value one less than the value where Y1=0	y-value one more than the value where Y1=0
$x^2-x=12$	$y=x^2-x-12$
$x^2+8x=9$	$y=x^2+8x-9$
$x^2=14x-24$	$y=x^2-14x+24$
$x^2+16x=-28$	$y=x^2+16x+28$

Calculation:

1.

• Let Y1 = $x^2-x-12$.
• Use the 2^nd [TABLE] option to get the table of values.
• Use the up and down arrow to locate the values of x where Y1 = 0.
• Identify the y -values above and below the value where Y1 = 0.
• Add these values in the table.

  

Equation	Related Function	Zeroes	y-value one less than the value where Y1=0	y-value one more than the value where Y1=0
$x^2-x=12$	$y=x^2-x-12$	-3	8	-6
		4	-6	8

2.

• Let Y1 = $x^2+8x-9$.
• Use the 2^nd [TABLE] option to get the table of values.
• Use the up and down arrow to locate the values of x where Y1 = 0.
• Identify the y -values above and below the value where Y1 = 0.
• Add these values in the table.

  

Equation	Related Function	Zeroes	y-value one less than the value where Y1=0	y-value one more than the value where Y1=0
$x^2-x=12$	$y=x^2-x-12$	-3	8	-6
		4	-6	8
$x^2+8x=9$	$y=x^2+8x-9$	-9	11	-9
		1	-9	11

3.

• Let Y1 = $x^2-14x+24$.
• Use the 2^nd [TABLE] option to get the table of values.
• Use the up and down arrow to locate the values of x where Y1 = 0.
• Identify the y -values above and below the value where Y1 = 0.
• Add these values in the table.

  

Equation	Related Function	Zeroes	y-value one less than the value where Y1=0	y-value one more than the value where Y1=0
$x^2-x=12$	$y=x^2-x-12$	-3	8	-6
		4	-6	8
$x^2+8x=9$	$y=x^2+8x-9$	-9	11	-9
		1	-9	11
$x^2=14x-24$	$y=x^2-14x+24$	2	11	-9
		12	-9	11

4.

• Let Y1 = $x^2+16x+28$.
• Use the 2^nd [TABLE] option to get the table of values.
• Use the up and down arrow to locate the values of x where Y1 = 0.
• Identify the y -values above and below the value where Y1 = 0.
• Add these values in the table.

  

Equation	Related Function	Zeroes( x-coordinate where Y1 = 0)	y-value one less than the value where Y1=0	y-value one more than the value where Y1=0
$x^2-x=12$	$y=x^2-x-12$	-3	8	-6
		4	-6	8
$x^2+8x=9$	$y=x^2+8x-9$	-9	11	-9
		1	-9	11
$x^2=14x-24$	$y=x^2-14x+24$	2	11	-9
		12	-9	11
$x^2+16x=-28$	$y=x^2+16x+28$	-14	13	-11
		-2	-11	13

(d)

To determine

To Examine:

The function values ( y-values) for x-values just before and just after a zero. What happens to the sign of the value before and after a zero?

Answer to Problem 67PPS

The function values have opposite signs just before and after the zero.

Explanation of Solution

Given Information:

From part (c) , we have the completed table :

Equation	Related Function	Zeroes( x-coordinate where Y1 = 0)	y-value one less than the value where Y1=0	y-value one more than the value where Y1=0
$x^2-x=12$	$y=x^2-x-12$	-3	8	-6
		4	-6	8
$x^2+8x=9$	$y=x^2+8x-9$	-9	11	-9
		1	-9	11
$x^2=14x-24$	$y=x^2-14x+24$	2	11	-9
		12	-9	11
$x^2+16x=-28$	$y=x^2+16x+28$	-14	13	-11
		-2	-11	13

Observation:

From the above table, for the zero x = -3 of function $y=x^2-x-12$, the y-values just before and after the zero are y = 8 and y = -6.

Similarly we can observe that the y-values for each zero changes sign just before the zero and after the zero.

So, the function values have opposite signs just before and after the zero.