1. 3.4 The linear function y = mx + b
Chapter: 3
Functions
3.4 The linear function y = mx + b

2. Real-world application of linear functions
Suppose that the port of Piraeus is located at point Ο (0, 0) of a Cartesian coordinate system of axes while a vessel follows a straight line course with the equation ε1: y = 2x – 4
How far east and south of the harbor will the vessel travel?
(Consider that the unit of the coordinates is set to miles).
Let us calculate the intercepts (points that the line crosses the horizontal and vertical axes):
x-intercept:
x’x: let y = 0
so 2x – 4 = 0
(=)2x = 4
thus, x = 2
The coordinates of the x-intercept are A(2,0).
y-intercept:
y’y: let x = 0
so y = 2·0 – 4
thus y = – 4
The coordinates of the y-intercept are με B(0, –4)
Therefore, the vessel travels 2 miles to east and 4 miles to the south of Piraeus port.

3. Try to draw the graph of the line ε1: y = 2x – 4 on a Cartesian coordinate system:
if x = 0, then y = 2. 0 – y = 0 – y = – 4
if y = 0, then = 2. x – = 2. x
2 = x
if x = 1, then y = 2. 1 – y = 2 – y = – 2 x 2 1 y -4 -2 ω
What does “m” stand for in the linear function y=mx+b;
m: the slope or gradient of the line
(m=tanω) b
What does “b” stand for?
b: the corresponding y value of the y-intercept

4. Let us explore the line equation y=mx+b using FluidMath!

5. Now try to solve the equation below for y and then illustrate it graphically (if you need help select a hint):

10. Draw the straight course lines of two vessels with equations y = 2 and x = 1 and check whether their paths cross at some point.
Every line of the form y = b is parallel to the x axis and intersects y axis at point (0, b).
Every line of the form x = c is parallel to the y axis and intersects the x axis at point (c, 0).
y = 2
K (1, 2)
Therefore, they meet at point K(1, 2).
x = 1

11. If at points Α(–1, 4) and Β(3, 2) exist buoys from fishing nets, verify whether the initial vessel will reach them at some point.
If point A lies on the line of the equation then its coordinates should satisfy y = 2x – 4.
For point Α: Let 4 = 2·(–1) – 4 or
4 = – 2 – 4
or 4 = – 6
which doesn’t hold. Therefore, the vessel doesn’t pass through point A.
For point Β: Let 2 = 2·3 – 4 or
2 = 6 – 4 or
2 = 2
which holds. Therfore, the vessel passes through point B.
Find the x-coordinate of the point through which the vessel passes and its y-coordinate is equal to –5.
If a given point belongs to line ε1, then its coordinates should verify the equation y = 2x – 4.
Hence – 5 = 2·x – 4 or
– = 2·x or
– 1 = 2·x or x = -1/2.

12. Solve equation ε2 for y in order to bring it in the form of y=mx+b:
Τhe courses of three different vessels are given by the equations ε2: 5y  5 = 10x, ε3: 5y = 15x and ε4: 4 = 2x – y. Examine their line courses relative to the initial vessel (ε1: y = 2x – 4).
Solve equation ε2 for y in order to bring it in the form of y=mx+b:
5y = 10x + 5
or y = 2x + 1
We observe that both lines ε1 and ε2 have the same slope value (m=2) thus, they move parallel to each other.
Solve the equation of the line ε3 for y similarly:
y = 3x
In this case, both lines ε1 and ε3 do not share the same value for slope (m1=2 and m3=3) therefore, they intersect at some point.
Solve the equation of the line ε4 for y similarly:
y = 2x  4
In this case, we observe that both lines ε1 και ε4 have the same value for slope (m=2) and same y intercept (b=-4) thus, they coincide. This means that they follow the exact same course.

13. Define a different line of equation that should cross the course of the initial vessel (y = 2x – 4, red line).
Define a different line of equation that should move parallel to the initial vessel (y = 2x – 4)

14. Derive the equation of the line of another vessel which moves parallel to the initial vessel (y = 2x – 4) and passes through the point D(5, 8).
Since the course of 1st vessel has the equation y = 2x – 4 and the 2nd vessel moves parallel to that, then the equation of the line of the 2nd vessel is of the form: y = 2x + β.
Since the 2nd vessel passes through the point D(5, 8), then its coordinates should satisfy the equation y= 2x + β.
Therefore, 8 = 25 + β or
8 = 10 + β or
8  10 = β or
β = -2.
As a result, the equation of the line of the 2nd vessel is equal to y = 2x – 2.

15. Consider another vessel with equation of line ε5: y = (λ  2)x + 4
Consider another vessel with equation of line ε5: y = (λ  2)x + 4. Calculate λ such that this vessel moves parallel to the initial one (ε1: y = 2x  4).
In order for ε1 // ε5, we need to set the two values for slope equal to each other (m1=m5):
λ  2 = 2 or
λ = or
λ = 4.
Consider another vessel with line of equation ε6: y = (2λ  3)x + 5. Calculate λ such that this vessel passes through the point F(5, 3).
Since the point Γ(5, 2) belongs to the equation of line ε6, the coordinates should satisfy the equation: y = (2 λ  3)x + 5.
Therefore, 2 = (2λ 3)(5) + 5 or
2 = 10λ or
10λ =  2
or 10λ = 18 or λ = 1,8.