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Equations and their Graphs
I. LINEAR EQUATIONS
A. GRAPHS
Any equation with first powers of x and/or y is referred to as a linear equation. When graphed, all ordered ( x , y )pairs that satisfy a linear equation form a straight line.
Example. Find 4 ordered pairs (including x and y intercepts) that satisfy 2 x + 3 y = 8. Graph the line.
2 ( 0 )+ 3 y = 8 y = 8 / 3 ⇒( 0 , 8 / 3 )
2 ( − 2 )+ 3 y = 8 y = 4 ⇒(− 2 , 4 )
2 x + 3 ( 0 )= 8 y = 4 ⇒( 4 , 0 )
2 x + 3 (− 2 )= 8 x = 7 ⇒( 7 ,− 2 )
Here are 3 more examples of graphs of linear equations.
(4/3, 0)
(4, −4) (0, −4)
(−6, −4)
(2, −2)
(2, 0)
x = 2
(7,−2)
(−2,4) (0,8/3)
(4,0)
x intercept
y intercept
y = 3 x − 4
y =− 4
B. SLOPE
The most important characteristic of a line is the value assigned to the ratio comparing the amount of vertical change to horizontal change. Known as slope , this value can be defined in many ways:
slope or x
y x x
y y x
y m ∆
2 1
2 1 run
rise changein
changein
In this course, slope will often be described as the Rate of Change and will (eventually) be associated with the term Derivative.
Example. Find the rate of change for each of the following lines.
a)
b)
Notice that the rate of change of a line does not depend on the points selected.
↑ delta means change
Rate of change 3
m =
− 3
8 1 ,
y 2 − y 1 =− 9
(^) − 3
2 4 ,
x 2 − x 1 = 6
∆ x = 5
∆ y = − ^10 3
2 x + 3 y = 6
Using (−1, 8/3) and (4, −2/3),
x
y m.
Using (−3, 4) and (6, −2),
2 1
x x
y y m.
Run = 3
Rise = 1
y = 3 x
Exercise 1. Which of the following tangent lines have
a) m > 0? b) m < 0? c) m = 0?
II. QUADRATIC EQUATIONS
Only linear equations have graphs that result in lines. The graphs of all nonlinear equations will be “curves”.
An equation in the form y = ax^2 + bx + c ( a ≠ 0 ), is referred to as “Quadratic” and its graph is a parabola. Here are examples of the graphs of two quadratic equations along with the tables used to find points on each.
Example. y = 4 − x^2
x y 0 4 1 3 − 1 3 ± 2 0 ± 3 − 5
A parabola that opens down is said to be “concave down”. The point (0, 4) is known as the vertex.
(0, 4)
(1, 3)
(2, 0)
(3, −5)
(−1, 3)
(−2, 0)
(−3, −5)
Answers
Example. y = x^2 − 4 x − 5
x y (^0) − 5 (^1) − 8 (^3) − 8 5 0 6 7 − 1 0
a
b x
Now substitute to find y.
y = 2 2 − 4 ( 2 )− 5 =− 9
We can also say that −9 is the minimum value of the equation. In our first example, 4 is the maximum value. Notice that a tangent line drawn at a minimum or maximum is horizontal.
Are you beginning to see a pattern? Horizontal tangent lines ⇒ slope 0 ⇒ maximum or minimum values ☺ ☺.
Exercise 2. Graph each parabola. Start by determining the vertex using a
b x 2
= −. Then find
2 points on each “side” of the vertex as well as x and y intercepts. State the maximum or minimum value.
a) y = x^2 − 3
b) y = x^2 − 4 x
c) y = 6 x − x^2
d) y = − x^2 + 6 x + 7
(5, 0)
(2, −9)
(−1, 0)
(0, −5)
This parabola is concave up with vertex at (2, −9). The vertex is so important that we’ll give you a formula to find it. Start by finding the x value using a
b x 2
Answers
Answers
Example. Here are two polynomials with notation describing end behavior.
The arrow is read "approaches". In calculus, this same end behavior will be expressed using slightly different notation along with the term "limit". We simply replace the term equation with function and the letter y with f ( x ). Here are the same graphs with their end behavior described using limit notation.
y
as x ,
y
as x , →−∞
y
as x ,
y
as x ,
f ( x ) = ∞ → ∞
limit f ( x ) x = −∞ →−∞
limit f ( x ) x
p ( x )
→∞
limit p ( x ) x = ∞ → −∞
limit p ( x ) x
- Increasing/Decreasing
Except when it remains constant (indicated by a horizontal line), an equation or function is either increasing or decreasing. Here is what we mean.
The ( a , b ) indicates an interval on x , not a point.
Let's look once again at our original graph along with all the information we've discussed.
- Turning Points: Maximum at x =− 3 and 6; Minimum at x = 2.
- End Behavior: =−∞ →−∞
limit f ( x ) x
→∞
limit f ( x ) x
- Increasing/Decreasing: f ( x ) is increasing on ( −∞, − 3 ) and ( 2 , 6 ) f ( x ) is decreasing on (− 3 , 2 )and ( 6 ,∞)
a (^) b a (^) b a b a b
The graphs above are said to be increasing on ( a , b ).
The graphs above are decreasing on ( a , b ).
− 3
2
6
f ( x )
A. VERTICAL AYMPTOTES
What happens with points "near" a value for which the function (equation) is undefined? Recall, division by 0 is undefined. In our first example, f ( x ) is undefined when x = 1. Let us examine points near this value.
There are 2 ways to describe this behavior.
Algebra version: as x → 1 +, f ( x )→∞ as x → 1 −, f ( x )→−∞
Calculus version: =∞ →^ +
limit ( ) 1
f x x = −∞ →−
limit ( ) 1
f x x
Every rational function with undefined values will have vertical asymptotes at these values. As the graph nears a vertical asymptote, it will "veer off" and head upward or downward depending on the direction of approach.
x 1
2 ( ) − = x f x
. 5
(. 5 )^2 =− −
f =
. 1
2 (. 9 ) =− − f =
. 01
2 (. 99 ) =− − f =
0
2 f ( 1 ) undefined
. 01
2
. 1
2
. 5
2
x approaching 1 from the left or x → 1 −
x approaching
1 from the right
or x → 1 +
x = 1
f ( x )
x
B. HORIZONTAL AYMPTOTES
To identify a horizontal asymptote we must examine end behavior. Once again a table will help.
Here's the end behavior described using limits:
limit ( )= 0 →∞
f x x
limit ( )= 0 → −∞
f x x
conclusion ⇒ y = 0 is a HA.
Can you see anything in our 2nd example, 1
x
x g x , that would lead you to conclude
there is a HA at y = 1? Examine the table below.
x g ( x )
10 11.
10 ≈
20 ≈
30 ≈
10 ≈ −
−
− 20 19 1.^05
20 ≈ −
−
30 ≈ −
−
limit ( ) 1 x
→ ∞
g x
limit ( ) 1 x
→ −∞
g x
conclusion
⇒ HA at^ y =^1
x 1
2 ( ) −
= x
f x
2 ≈
2 ≈
2 ≈
2 ≈− −
− 20 21.^09
(^2) ≈− −
− 30 31 -.
2 ≈ −
x →∞
x →∞
f ( x )→ 0
f ( x )→ 0
10
− 10
f ( x )
x
To find Horizontal Aymptotes examine end behavior, i.e., find the limit as x →∞or
Answers to Exercises for Equations and their Graphs
Exercise 1.
a) m > 0 for l 1 and l 5
b) m < 0 for l 3
c) m = 0 for l 2 and l 4
Exercise 2.
a) y = x^2 − 3
x y (^0) − 3 ± 1 − 2 ± 3 6
*To find x intercepts, solve x^2 − 3 = 0
b) y = x^2 − 4 x
2 2 ( 1 )
x =
x y (^2) − 4 (^1) − 3 0 0 4 0 5 5
vertex and y -intercept
x -intercept
( 3 ,0)
(0, −3)
(− 3 , 0)
(−1, −2)
(−3, 6) • (3, 6)
(1, (^) −2)
vertex
intercepts
(2, −4)
(4, 0)
Exercise 2. (continued)
c) y = 6 x − x^2
3 2 ( 1 )
x =
x y 3 9 4 8 6 0 0 0 1 5
d) y = − x^2 + 6 x + 7
3 2 ( 1 )
x =−
x y 3 − 9 + 18 + 7 = 16 0 7 7 0 − 1 0 6 7
vertex
intercepts
This parabola is concave down and has a maximum value of 9 when x = 3.
(3, 16)
vertex y -int x -int
To find x -intercept, solve
2
2
x x
x x
x x
x x
(3, 9)
(4, 8)
(6, 0)
