graphs of first degree functions

1. a.2x – 3y = 12 b. –3y = 12 c. 2x = 12
If both x and y are
present, we get a
tilted line.
If the equation is
y = c, we get a
horizontal line.
The graphs of the equations Ax + By = C are straight
lines. It's easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get the y–intercept, and set
y = 0 for the x–intercept. If there is only one variable
in the equation, we get a vertical or a horizontal line.
(6,0)
(0,–4) y = –4
x = 6
If the equation is
x = c, we get a
vertical line.
First Degree Functions

2. Linear Equations and Lines
Example A. A river floods regularly, and on a rock by
the river there is a mark indicating the highest point
the water level ever recorded.
At 12 pm July 11, the water level is 28 inches below
this mark. At 8 am July 12 the water is 18 inches
below this mark. Let x = time,
y = distance between the water level and the mark.
Find the linear function y = f(x) = mx + b
of the distance y in terms of time x.
Since how the time was measured is not specified,
we may select the stating time 0 to be time of the
first observation.
By setting x = 0 (hr) at 12 pm July 11,
then x = 20 at 8 am of July 12.

3. Using the point (0, 28) and the point–slope formula,
y = – ½ (x – 0) + 28 or
Δy
Δx
28 – 18
0 – 20
Equations of Lines
In particular, we are given that at x = 0 →y = 28,
and at x = 20 → y = 18 and that we want the
equation y = m(x – x1) + y1 of the line that contains
the points (0, 28) and (20, 18).
=The slope m = = –1/2
– xy = + 28
2
The linear equation that we found is also called the
trend line. So if at 4 pm July 12, i.e. when x = 28,
we measured that y = 12” but based on the formula
prediction that y should be – 28/2 + 28 = 14”, we may
conclude that the flood is intensifying.

4. More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B. Find the equation of the line L that
passes through (2, –4) and is perpendicular to
4x – 3y = 5.
Solve for y to find the slope of 4x – 3y = 5
4x – 5 = 3y
4x/3 – 5/3 = y
Hence the slope of 4x – 2y = 5 is 4/3.
Therefore L has slope –3/4. So the equation of L is
First Degree Functions
y = (–3/4)(x – 2) + (–4) or y = –3x/4 – 5/2.