chapter1_part2.pdf

1. Chapter 1, Part 2
Chapter 1, Part 2
Linear Equations
Linear Equations
Q
Quadratic equation
uadratic equation
Exponential and
Exponential and
Logarithmic Equations
Logarithmic Equations

2. Writing Equations and Graphing
●
These activities introduce rates of change and
defines slope of a line as the ratio of the vertical
change to the horizontal change.
●
This leads to graphing a linear equation and writing
the equation of a line in three different forms.

3. The equation
y = 2x + 6
is a linear equation
because it is the
graph of a straight
line and each time x
increases by 1 unit, y
increases by 2
X Y = 2x + 6 Y (x, y)
1 2(1) + 6 8 (1, 8)
2 2(2) + 6 10 (2, 10)
3 2(3) + 6 12 (3, 12)
4 2(4) + 6 14 (4, 14)
5 2(5) + 6 16 (5, 16)

4. Remember, linear equations have constant
slope. For a line on the coordinate plane, slope
is the following ratio. This ratio is often referred
to as “rise over run”.

5. x-intercept – the x-coordinate of the point where
the graph of a line crosses the x-axis (where y =
0).
y-intercept – the y-coordinate of the point where
the graph of a line crosses the y-axis (where x =
0).
Slope-intercept form (of an equation) – a linear
equation written in the form y = mx +b, where m
represents slope and b represents the y-intercept.
Standard form (of an equation) – an equation
written in the form of Ax + By = C, where A, B, and
C are real numbers, and A and B are both ≠ 0.

6. Find the x-intercept and y-intercept of each line.
Use the intercepts to graph the equation.
1) x – y = 5
2) 2x + 3y = 12
3) 4x = 12 + 3y
4) 2x + y = 7
5) 2y = 20 – 4x

7. Slope-intercept
Form
●
An equation
whose graph is a
straight line is a
linear equation.
Since a function
rule is an
equation, a
function can also
be linear.
• m = slope
• b = y-intercept
y = mx + b
(if you know the slope and where
the line crosses the y-axis, use
this form)

8. For example in the equation;
y = 3x + 6
m = 3, so the slope is 3
b = +6, so the y-intercept is +6
Let’s look at another:
y = 4/5x -7
m = 4/5, so the slope is 4/5
b = -7, so the y-intercept is -7
Please note that in the slope-intercept formula;
y = mx + b
the “y” term is all by itself on the left side of the equation.
That is very important!

9. Using slope-intercept form to
find slopes and
y-intercepts
The graph at the right shows
the equation of a line both in
standard form and slope-
intercept form.
You must rewrite the
equation 6x – 3y = 12 in
slope-intercept to be able to
identify the slope and y-
intercept.

10. Write the equation of a line in slope-intercept
form that passes through points (3, -4) and
(-1, 4).
1) Find the
slope.
4 – (-4) 8
-1 – 3 -4
m = -2
2) Choose either point and
substitute. Solve for b.
y = mx + b (3, -4)
-4 = (-2)(3) + b
-4 = -6 + b
2 = b
Substitute m and b in equation.
y = mx + b
y = -2x + 2
Do you
remember
the slope
formula?

11. Write the equation of the line in
slope-intercept form that passes through each
pair of points.
1) (-1, -6) and (2, 6)
2) (0, 5) and (3, 1)
3) (3, 5) and (6, 6)
4) (0, -7) and (4, 25)
5) (-1, 1) and (3, -3)

12. There are several ways to graph a straight line
given its equation.
Let’s quickly refresh our memories on equations of straight
lines:
Slope-intercept Point-slope Horizontal line Vertical line
y = mx + b
W hen stated in “y=”
f or m, it quickly gives
the slope, m, and
wher e the line
cr osses the y-axis, b,
called the y-
int er cept.
y - y1 = m(x –x1)
when gr aphing, put
this equation into
“y=” f or m to easily
r ead gr aphing
inf or mation.
Y = 3 (or any # )
Hor izontal lines have
a slope of zer o –
they have “r un”, but
no “r ise” –all of the
y values ar e 3.
X = -2 (or any # )
Ver tical line have no
slope (it does not
exist) –they have
“r ise”, but no “r un” –
all of the x values
ar e -2.

13. Practice with Equations of Lines
Answer the following questions dealing with equations
and graphs of straight lines.
1) Which of the following equations passes through
the points (2, 1) and (5, -2)?
a. y = 3/7x + 5 b. y = -x + 3
c. y = -x + 2 d. y = -1/3x + 3

14. E.g. 2—Solving an Equation Involving
Fractions
Solve the equation
 
2 3
6 3 4
x
x
 
 
 
 
 



2 3
12 12
6 3 4
2 8 9
8 7
8
7
x
x
x x
x
x

15. Equation Involving Fractional Expressions
Solve the equation

 
   
2
1 1 3
1 2 2
x
x x x x

   
     
   
   
   
    
  

2
1 1 3
( 1)( 2) ( 1)( 2)
1 2 2
( 2) ( 1) 3
2 1 3
4
x
x x x x
x x x x
x x x
x x
x

16. Now, we try to substitute x = 4
back into the original equation.
– To check that x=4 is a solution
– So x=4 is a solution.

18. 
b
a
2
1. The standard form of a quadratic equation is y = ax2
+ bx + c.
2. The graph of a quadratic equation is a parabola.
3. When a is positive, the graph opens up.
4. When a is negative, the graph opens down.
5. Every parabola has a vertex. For graphs opening up, the vertex
is a minimum (low point). For graphs opening down, the vertex
is a maximum (high point).
6. The x-coordinate of the vertex is equal to .

19. To find the x coordinate of the vertex, use the equation
Then substitute the value of x back into the equation of the parabola
and solve for y.
x
b
a
 
2
You are given the equation y=-x2
+ 4x –1. Find the coordinates
of the vertex.
x
b
a
 
2
x  2

 
a b
1 4
,
( 
x  
4
2 1
)( )
2
y x x
   
4 1
y    
4 8 1
y 3
y    –
2
2 4 2 1
( ) ( )
The coordinates of the vertex are (2,3)
Substitute and solve for y

20. Choose two values of x that are to the right or left of the x-coordinate of the vertex.
Substitute those values in the equation and solve for y.
Graph the points. (Keep in mind the value of a as this will help you determine which way
the graph opens.)
Since a parabola is symmetric about the vertical line through the vertex, you can plot
mirror image points with the same y-values on the “other side” of the parabola.
x y = -x2
+ 4x – 1 y
1
-1
y = -(1)2
+ 4(1) – 1
y = -1 +4 – 1
2
y = -(-1)2
+ 4(-1) – 1
y = -1 – 4 – 1
-6

21. x
y
Plot the vertex and the points from your
table of values: (2,3), (1,2), (-1,-6).
Use the symmetry of parabolas to plot
two more points on the “other side” of
the graph. The point (1,2) is one unit
away from the line of symmetry, so we
can also plot the point (3,2). The point
(-1,-6) is three units away from the line
of symmetry, so we can also plot the
point (5,-6).
Sketch in the parabola.

22. Find the vertex of the following quadratic equations. Make a table of
values and graph the parabola.
1 4
2
. y x x
 
2 2 3
2
. y x
  
3 6 4
2
. y = x x
 

23. y x x
 
2
4
a b
  
1 4
x
b
a
x
x
 
 


2
4
2 1
2
( )
y x x
y
y
y
 
 
 
 
2
2
4
2 4 2
4 8
4
( )
The vertex is at (2,-4)
x y = x2
-4x y
1 y=12 - 4(1) -3
0 y = 02 - 4(0) 0
y
x
Notice, a is positive, so the graph opens up.

24. y x
  
2 3
2
a b
  
2 0
x
b
a
x
x
 
 


2
0
2 2
0
( )
y x
y
y
  
  

2 3
2 0 3
3
2
2
( )
The vertex is at (0,3)
x y = -2x2
+ 3 y
-1 y = -2(-1)2
+ 3 1
-2 y = -2(-2)2 + 3 -5
x
y
Notice, a is negative, so the graph opens
down.

25. Solving Quadratic Equations Using the
Quadratic Formula
1. Solve quadratic equations using the quadratic formula.
2. Use the discriminant to determine the number of real
solutions that a quadratic equation has.
3. Find the x- and y-intercepts of a quadratic function.
4. Solve applications using the quadratic formula.

26. Quadratic Formula
To solve ax2
+ bx + c = 0, where a 0, use

2
4
2
b b ac
x
a
  


27. 2
4
2
b b ac
x
a
  

a = 2
      
 
2
7 7 2 3
4
2
2
x
  



7 49 24
4
x
 

7 25
4
x


7 5
4
x


7 5
4
x


7 5
4
x


3
x 
1
2
x 
Solve:
Solve: 3
7
2 2

 x
x
0
3
7
2 2


 x
x






3
,
2
1
2 real rational solutions
, b = –7 , c = 3

28. 2
2 5 6
x x
  
      
 
2
2 2 4 1 11
2 1
x
     

2 4 44
2
x
 

2 48
2
x


2
2 11 0
x x
  
a = 1, b = –2, c = –11.
2 4 3
2 2
x  
1 2 3
x  
Solve:
Solve:
 
3
2
1
,
3
2
1 

2 real irrational solutions
16 ∙3
2 4 3
2
x



29. 2
2
4 i
x


Solve:
Solve:
2 non-real complex solutions
x
x 4
5
2


0
5
4
2


 x
x
a = 1, b = -4, c = 5
      
 
1
2
5
1
4
4
4
2






x
2
4
4 


x
 
i
i 
 2
,
2
i
x 
 2
2 1
1

30. Solve using the quadratic formula.
a)
b)
c)
d)
2
3 8 2 0
x x
  
8 2 10
6

8 10
3

4 10
3
 
2 22


31. 7 25
4
x


0
3
7
2 2


 x
x






3
,
2
1
2 real rational
solutions
2 48
2
x


2
2 11 0
x x
  
 
3
2
1
,
3
2
1 

2 real irrational
solutions
2 non-real
complex solutions
0
5
4
2


 x
x
2
4
4 


x
 
i
i 
 2
,
2
What made the difference? The Discriminant
The Discriminant
ac
b 4
2


32. Discriminant:
The discriminant is the radicand, b2
– 4ac, in the
quadratic formula.
The discriminant is used to determine the number and
type of solutions to a quadratic equation.

33. Use the discriminant to determine the number and type of solutions.
Use the discriminant to determine the number and type of solutions.
2
2 5 1
x x
 
2
2 5 1 0
x x
  
Evaluate the discriminant: b2
– 4ac.
    
2
5 4 2 1

a = 2, b = 5, c = 1
25 8
 
17

Two real irrational solutions.
Positive but not a perfect square.
Discriminant:

34. Determine the number and type of solutions.
a) Two real rational solutions
b) Two real irrational solutions
c) One real rational solution.
d) Two non-real complex solutions.
2
7 6
x x
 

35.   12
2


 x
x
x
f
12
2


 x
x
y
12
0 2


 x
x
What are we finding?
x-intercepts
  
3
4
3
4
0






x
x
x
x
 
12
,
0 
y-intercept
      12
12
0
0
0
2





f
   
0
,
3
0
,
4


36. Exponential and Logarithmic
Equations
Essential Questions:
How do we solve
exponential and
logarithmic
equations?

37. E.g. 4—Graphing a Logarithmic Function by Plotting Points
Sketch the graph of
f(x) = log2x
• To make a table of values,
we choose the x-values to be
powers of 2 so that we can
easily find their logarithms.

38. We plot these points and connect them
with a smooth curve.
E.g. 4—Graphing a Logarithmic Function by Plotting Points

39. Graphs of Logarithmic Functions
The figure shows the graphs of the family
of logarithmic functions with bases 2, 3, 5,
and 10.

40. Find the domain of f(x)= log10 x
Answer {x| x > 0}=(0,∞).
• Find the domain of h(x) = log10(x – 3)
is:
{x | x – 3 > 0} = {x | x > 3} = (3, ∞)

41. Natural Logarithms
If we substitute a = e and write “ln”
for “loge” in the properties of logarithms,
we obtain the following properties
of natural logarithms.

42. Properties of Natural Logarithms
Property Reason
1 ln 1 = 0 We must raise e to the power 0
to get 1.
2 ln e = 1 We must raise e to the power 1
to get e.
3 ln ex
= x We must raise e to the power x
to get ex
.
4 eln x
= x ln x is the power to which e must
be raised to get x.

43. Find the domain of the function
f(x) = ln(4 – x2
)
• As with any logarithmic function, ln x is defined
when x > 0.
• Thus, the domain of f is:
     
 
 
2 2
| 4 0 | 4 | 2
| 2 2
2, 2
x x x x x x
x x
     
   
 

44. Graphing Exponential
Functions
Objectives:
1. Graph exponential growth functions.
2. Graph exponential decay functions.

47. Example 1
Graph y = 3x
.
x y
-3
-2
-1
0
1
2
Complete the table.
Substitute -3 for x
y = 3-3
y = 1/27 or 0.04
Write 0.04 in the table
0.04
Substitute -2 for x
y = 3-2
y = 1/9 or 0.1
0.1
0.3
1
3
9

48. x y
-2
-1
0
1
2
Complete the table.
Substitute -2 for x
y = 9
9
3
1
0.3
0.1

49. Graphs of Exponential Functions
Graph of f(x) = ax
, a > 1 Graph of f(x) = a-x
,
a > 1
Domain ( , , ) Domain:( , )
Range :(0 , ) Range :(0 , )
Intercept :(0 ,1) Intercept :(0 ,1)
Increasing on :( , ) Increasing on :( ,

50. Solving Simple Equations
Original Rewritten Solution
Equation Equation
5
a. 2 32 2 2 5
x x
x
  
b. ln ln3 0 ln ln3 3
x x x
   
2
1
c. 9 3 3 2
3
x
x
x

 
  
 
 
d. 7 ln ln 7 ln 7
x x
e e x
  
ln 3 3
e. ln 3 x
x e e x e
 
  
10
log 1 1
10
1
f. log 1 10 10 10
10
x
x x
 
   

51. Solving Exponential Equations
●
Solve each equation
a. b.
72
x
e  3(2 ) 42
x

Solution:
a. 72
x
e 
ln ln 72
x
e  Take natural log of
each side.
ln 72
x 
2 14
x

2 2
log 2 log 14
x

2
log 14
x 

52. Solving an Exponential Equation
●
Solve and approximate the result to three decimal
places.
5 60
x
e  
Solution:
5 60
x
e   Original equation.
55
x
e  Subtract 5 from each side.
ln ln55
x
e  Take natural log of each side.
ln55
x  Inverse Property.

53. Solve
2 5
2(3 ) 4 11
n
 
2 5
2(3 ) 4 11
n
 
2 5
2(3 ) 15
n

2 5 15
3
2
n

2 5
3 3
15
log 3 log
2
n

3
15
2 5 log
2
n  
3
2 5 log 7.5
n  
3
5 1
log 7.5
2 2
n  

54. Solving a Logarithmic Equation
●
Solve 5
2log 3 4.
x 
Solution: 5
2log 3 4
x 
5
log 3 2
x 
5
log 3 2
5 5
x

3 25
x 
25
3
x 

55. Checking for Extraneous Solutions
●
Solve 10 10
log 5 log ( 1) 2
x x
  
Solution:
10 10
log 5 log ( 1) 2
x x
  
10
log [5 ( 1)] 2
x x  
2
10
log (5 5 ) 2
10 10
x x


2
5 5 100
x x
 
2
5( 20) 0
x x
  
( 4)( 5) 0
x x
  
4 5
x x
 
The solutions appear to be
5 and -4. However, when
you check these in the
original equation, only x =
5 works.