Chapter 2 - Examples of Problem in Homework 3 and Workshop Examples
Linear Functions

Example 1. True or False? - Answer Key

(a) The slope of two perpendicular lines are the same.

(b) The rate of change of two parallel lines are the same.

(c) If the slope of a line L(1) perpendicular to another line, L(2) is 4 then the slope of L(2) is -¼?

(d) The slope of a line perpendicular to a line with this formula: y = -2x + 6 is  -2?

(e) The intersection of  two parallel lines always intersects at the same point.


Example 2.  Say which of the following functions is/are linear from the table below:

x
 g(x)
 h(x)
1
 10
 100
3
 20
 80
5
 40
 60
7
 80
 40
9
 160
 20

Answer: h(x)  is a linear function since for the same change in x of 2 there is a constant change in h of -20.

Example 3. Find the points of intersection of the following functions: g(x) = 2x + 2 and h(x) = -3x -3

Answer:  (-1, 0)  : Set g(x) = h(x),  2x + 2 = -3x - 3, 5x = -5, x = -1 and when x =  -1, the function equal 2(-1) + 2 = 0
                    So x = -1 and y = 0.

Example 4. Find a possible formula of the function in example 2 for h(x).

Answer: h(x) = -10x + 110

Check to see that function is linear: for each change in x of 2, there is the same change in h(x) of -20, so linear with
    possibly formula of  h(x) = mx + b.

    First find the slope, m: m = change in h(x) divided by change in x: Lets take the first 2 points to solve for m:
        m = (80 - 100) / (3 - 1) = -20 / 2 = -10; So formula so far is h(x) = -10x + b,
        to find b: use any point say first point in table, x = 1 and h(x) = 100:   100 = -10(1) + b
        Solve for b = 100 + 10 = 110, So formula is h(x) = -10x + 110
       
Example 5. Find the formula for a linear function with these two points: (25, 125) and (15, 100):

Answer: y= 2.5x + 62.5

First order the points in increase order of x: (15, 100), (25, 125)
    Note: (x1, y1) and (x2, y2) respectively.

Since linear, y = mx + b, First find m = change in y / change in x:  m = (y2 - y1) / (x2 - x1) = (125 - 100) / (25 - 15) = 25 / 10 = 2.5

So formula so far is y = 2.5x + b, and using first point (15, 100) to find b we get: 100 = 2.5(15) + b, b = 100 - 37.5 = 62.5

Formula is y = 2.5x + 62.5

Example 6. Find the formula for the linear depreciation of the value, V of a car from $35,000 to $25,000
over a 3 year time, t h period.

Answer:  V= -3333.33t + 35000

First extract the points from the word problem: (t, V) => (0, 35000) and (3, 25000).

Since linear depreciation, V = mt + b where b = 35000 since t = 0 and m is negative since a decreasing function.
To find the slope, m = change in V / change in t = (V2 - V1) / (t2 - t1) = (25000 - 35000) / (3 - 0) = -10000/3 = -3333.33

So formula is V = -3333.33t + 35000

Example 1 - Answer Key: (a) False  (b) True  (c) True   (d) False, it is ½   (e) False