Section 18-1: Graphical Representation of Linear Equations and
Functions
Learning Outcome 1
Prepare a table of solutions and locate the solutions on a coordinate system: f(x) = 2x – 5
Learning Outcome 2
Write x + 3 = 5 as a function and graph the function.
x+3=5
Set equation equal to zero.
x–2=0
Write in function notation.
f(x) = x – 2
Assign x values and find f(x).
f(0) = 0 – 2 = –2
Let x = 0
f(2) = 2 – 2 = 0
Let x = 2
f(4) = 4 – 2 = 2
Let x = 4
Learning Outcome 3
A solution for the equation 2x – 3y = 6 is (6, 2). Verify the solution.
2x – 3y = 6
Substitute 6 for x and 2 for y.
2(6) – 3(2) = 6
12 – 6 = 6
6=6
True
Learning Outcome 4
Solve 3x + y = 7
3x + y = 7
3x + 16 = 7
3x = 7 – 16
Substitute 16 for y.
Sort terms.
Combine terms.
3x = –9
x = –3
Divide.
Section 18-2: Graphing Linear Equations with Two Variables Using
Alternative Methods
Learning Outcome 1
Graph by using the intercepts method: 2x – 5y = 10
2x – 5y = 10
To find x-intercept, let y = 0.
2x – 5(0) = 10
2x = 10
The coordinate pair is (5, 0).
x=5
2x – 5y = 10
To find the y-intercept, let x = 0.
2(0) – 5y = 10
–5y = 10
y = –2
The coordinate pair is (0, –2).
Locate the two intercepts using the coordinate pairs.
Draw the graph through the two points.
Learning Outcome 2
2
x−4
3
The slope-intercept form of the linear equation is y = mx + b where m = slope and b = y-coordinate
of the y-intercept.
2
2
y = x−4
m = ; b = –4
3
3
Use the slope-intercept method to graph: y =
2
, and
3
begin at the point (0, –4) and count up 2 (rise) and right 3 (run) to
locate the second point. The coordinates of this second point are
(4, –2). Draw the line through the two points..
First, locate the y-intercept, (0, –4). Then use the slope,
Learning Outcome 3
Use your graphing calculator to graph the equation y = –3x + 2. Before you graph the equation,
examine it and predict where it will cross the y-axis and describe its slope. From the equation it is
determined the line will cross the y-axis 2 units above the x-axis. Because the equation has a
negative slope, it might be expected to slope upward to the left.
Learning Outcome 4
Write the equation as a function, graph the function and find the solution from the graph.
3x − 5 = 7
3 x − 12 = 0
f ( x ) = 3 x − 12
Set the equation equal to zero.
Write in function notation.
f (−5) = 3(−5) − 12
= −15 − 12
= −27
f (0) = 3(0) − 12
= 0 − 12
= −12
f (4) = 3(4) − 12
= 12 − 12
=0
f (6) = 3(6) − 12
= 18 − 12
=6
The line crosses the x axis at x = 4. Thus the solution is x = 4.
Section 18-3: Graphing Linear Inequalities with Two Variables
Learning Outcome 1
Graph the solution set for y < –3x + 4. Graph the equation y = –3x + 4, which is the boundary. Use
−3
the slope-intercept method. The y-intercept is (0, 4) and the slope is
. The line is dashed to
1
indicate that it is not included in the solution set.
Test (0, 0) to see if it makes a true statement with the inequality. This determines which side of the
boundary is shaded as the solution set.
y < –3x + 4
0 < –3(0) + 4 ?
0<4
True
Therefore, shade the side of boundary that includes the point (0, 0) (see graph).
Section 18-4: Graphing Quadratic Equations and Inequalities
Learning Outcome 1
Identify the linear and quadratic equations from the following.
(a) 5x2 – 3x = 10
(c) x – y = 8
(b) 2xy – 3 = y
(d) x = y3 – 2x + 3
(a) quadratic equation
(c) linear equation
(b) quadratic equation
(d) cubic equation
2xy has degree 2.
Prepare a table of solutions of five ordered pairs. Plot the points on a rectangular coordinate
system. Connect the points with a smooth curve.
y − x2 + 6x + 8
Learning Outcome 2
Graph: y = x2 – 4x – 12
b
−4
−4
x−
=−
=−
=+2
2a
2(1)
2
Axis of symmetry = x = −
Find the x-coordinate of the vertex by using −
b
.
2a
b
.
2a
x=2
y = (+2)2
y = 4 – 8 – 12
y = –16
(+2, –16)
Find the x-intercepts.
0 = x2 – 4x – 12
0 = (x – 6) (x + 2)
x–6=0 x+2=0
x=6
x = –2
4(+2) – 12
Find y-coordinate of the vertex by substituting.
Coordinates of the vertex.
Substitute y = 0.
Sketch graph.
Learning Outcome 3
Examine the graph of the equation y = x2 – 2x –35 to find the solutions.
The graph crosses the x-axis at (6,0) and (–4, 0).
The solutions are x = –4 and x = 6.
Learning Outcome 4
Use a calculator to graph the equation: y ≥ x2 – 3x – 10 .Clear and initialize the calculator.
Section 18-5: Graphing Other Nonlinear Equations
Learning Outcome 1
Prepare a table of solutions and graph the function f(x) = 4x
Learning Outcome 2
Use the vertical line test to determine which graph is the graph of a relation and which is the graph
of a function.
(a)
(b)
(a) is a relation and not a function because a vertical line intersects the graph in 2 points.
(b) is a function because any vertical line intersects the graph in only one point.
Learning Outcome 3
Determine the domain and range of the relation.
The domain is all values on the graph for the
independent variable (x-values). [–6, 6]
The range is all values on the graph for the
dependent variable (y-values). [–3, 3]