Advance algebra

A powerpoint compilation
MIDTERM LESSONS IN
ADVANCE ALGEBRA

WORD PROBLEMS
The following steps must be followed in
order to solve word problems systematically.
Read and understand the problem.
1.Draw a diagram or table if possible.
2. Represent the unknown with a variable.
3. Write the other quantities in terms of that
variable.

4. Write the equation that models the
situation in the problem.
5. Solve the equation.
6. Answer the question in the problem.
7. Check the answer by using it in solving the
original problem.

WORD PROBLEMS EXAMPLES
NUMBER
PROBLEMS
MIXTURE
PROBLEMS
RATE, TIME
& DISTANCE
PROBLEMS
WORK
PROBLEMS
COIN
PROBLEMS
GEOMETRIC
PROBLEMS

The sum of two numbers is 99. The second is 3
more than the first. What are the numbers?
Solution:
Let x = first number
X = 3 = second number
Equation:
x + x + 3 = 99
2x + 3 = 99
2x = 99 – 3
2x = 96
x = 96/2
x = 48
Check:
48 + 51 = 99
99 = 99

How many grams of 8% acid solution and how
many grams of 15% acid solution must be mixed
to obtain 10 gr4ams of a 12% acid solutions?
8% acid
solution
8% x 0.08x
15% acid
solution
15% 10-x 0.15(10-x)
Mixture 12% 10 0.12(10)

Equation:
0.08x + 0.015(10-x) = 0.12(10)
0.08 + 1.5 – 0.15x = 1.2
0.08x – 0.15x = 1.2-1.5
-0.07x = -0.30
x = -0.30/-0.07
x = 4.49 grams

RATE, TIME &
DISTANCE
PROBLEMS

John drove 4,000 meters in 5 minutes. Part of the
trip was at 10 m/s and the r4est at 15 m/s. Find
the time spent for the 10 m/s speed.
Solution:
Time= distance traveled/ speed (T= d/s)
t= time
d= distance
s= speed
Let t1= time spent with speed of 10 m/s
5min- t1 or 300s – t1 = time spent with speed
of 15 m/s

Total distance traveled = 4,000 meters;
D1 = distance traveled with lower speed
D2 = distance traveled with higher speed
Equation:
Total distance = d1 +n d2
4,000 = (10m/s)(t1) + (15m/s)(300s – t1)
4000 = 10t1 + 4500 -15t1
-10t1 + 15t1 = 4500 – 4000
5t1 = 500
t1= 500/5
t1 = 100 seconds or 1 min and 40 seconds

George can paint a house in 8 hours, John in 10
hours, and Paul in 12 hours. How long will it take
to do the job if George and John work 2 hours and
then George and Paul finish the job.
George 1/8 x/8
John 1/10 2/10
Paul 1/12 (x-2)/12
Solution:

Equation:
x/8 + 2/10 + (x-2)/12 = 1
120(x/8) + 120(2/10) + 120(x-2)/12 = 120 (Multiplying by
the led (120))
15x + 24 + 10x – 20 = 120
25x + 4 = 120 - 4
25x = 116
x = 116/25
or 4 and 16/25 hours

Jason has 50 coins, all in 5 and 1 peso coins,
amounting to Php 130.00. How many 5 peso coin
does he have?
Solution:
Let x = number of 5 peso coins
Then 50-x = number of 1 peso coins
5x + 1(50-x) = 130
5x + 50 - x = 130
4x + 50 = 130
4x = 80
x = 20
He has 20 5-peso coins and 30
1-peso coins.
(50-x = 50-20 = 30)

If one side of a square is doubled in length and
the adjacent side is decreased by two centimeters,
the area of the resulting rectangle is 96 square
centimeters larger than that of the original
square. Find the dimensions of the rectangle.

2x2 – 4x = x2 + 96
x2 – 4x – 96 = 0
(x – 12)(x + 8) = 0
x = 12 or x = –8
Solution:
Square’s side length: x
One side is doubled: 2x
Next side is decreased by two: x – 2
Square’s area: x2
Rectangle’s area: (2x)(x – 2) = 2x2 – 4x
New area is 96 more than old area: 2x2 – 4x = x2 + 96

SOLVING QUADRATIC EQUATIONS
SQUARE
ROOT
PROPERTY
COMPLETING
THE
SQUARE
QUADRATIC
FORMULA
DISCRIMINANT FACTORING

The square root property is one
method that is used to find the
solutions to a quadratic (second
degree) equation. This method
involves taking the square roots of
both sides of the equation.

Steps:
1. Before taking the square root of each side, you
must isolate the term that contains the squared
variable.
2. Once this squared-variable term is fully isolated,
you will take the square root of both sides and
solve for the variable.

3. We now introduce the possibility of two roots for
every square root, one positive and one negative.
4. Place a sign in front of the side containing the
constant before you take the square root of that
side.

Example:
1.x2
= 12
𝑥2 = ±√12
𝑥2 = 12
𝑥 = 4(3) = ±2 3
… the squared-variable term is isolated,
so we will take the square root of each
side
… notice the use of the sign, this will
give us both a positive and a negative
root
… simplify both sides of the equation,
here x is isolated so we have solved
this equation
So, the values of 𝑥 𝑎𝑟𝑒 2 3𝑎𝑛𝑑 − 2√3.
Solution:

An alternate method of solving quadratic
equations is called completing the
square. Actually completing the square
involves using the square root
property. However, we have a few steps to
setting up the problem before we can use
completing the square.

Example:
1. 𝑥2
− 2𝑥 = 8
Solution:
𝑥2 − 2𝑥 = 8 … first isolate both terms containing the
variable onto one side of the equation
𝑥2
− 2𝑥 + 1 = 8 + 1 … next take of the middle term
now square this number and
add to both sides

(𝑥 − 1)2= 9 … the right-hand side will factor
into the square of a binomial
(𝑥 − 1)2=± 9 … now use the square root
property and solve
𝑥 − 1 = ±3
𝑥 = 1 ± 3
𝒙 = 𝟏 + 𝟑 = 𝟒 𝑎𝑛𝑑
𝒙 = 𝟏 − 𝟑 = −𝟐

In elementary algebra, the quadratic
formula is the solution of the quadratic
equation. There are other ways to solve the
quadratic equation instead of using the
quadratic formula, such as
factoring, completing the square, or graphing.
Using the quadratic formula is often the
most convenient way.

Example:
Solution:
By Quadratic Formula: a = 1, b = 2, c = -8
Substitute the values
and solve.

The discriminant is the expression b2 – 4ac. The value
of the discriminant can be used to determine the
number and type of roots of a quadratic equation.
We used the discriminant to determine whether a
quadratic polynomial could be factored. If the value of
the discriminant for a quadratic polynomial is a perfect
square, the polynomial can be factored. The
Discriminant can be negative, positive or zero.

REMEMBER:
If b2 – 4ac > 0 the equation will have two distinct
real roots.
If b2 – 4ac = 0 the equation will have repeated
roots.
If b2 – 4ac < 0 the equation will have no real
roots.

1. 9x2+6x+1=0
a=9, b=6, c=1
b2-4ac=(6)2-4(9)(1)
=36-36=0
1 real solution
Examples:
(Just substitute the values of a, b, and c.)
2. 9x2+6x-4=0
a=9, b=6, c=-4
b2-4ac=(6)2-4(9)(-4)
=36+144=180
2 real solutions

We use the Distributive Property to expand
Algebraic Expressions. We sometimes need to
reverse this process by factoring, an
expression as a product of simpler ones.
Example:
𝒙 𝟐
-4= (x-2)(x+2)

A. FACTORING BY COMMON FACTORS
Example:
1. 𝟑𝒙 𝟐
-6x The greatest common factor of
the terms 𝟑𝒙 𝟐
-6x is 3x, so we have
𝟑𝒙 𝟐
-6x= 3x(x-2)

B.FACTORING TRINOMIALS
To factor trinomial of the form 𝐱 𝟐
+ 𝐛 𝐱
+𝐜,
we note that (x+r)(x+s)= 𝒙 𝟐
+(r+s)x+rs,
so we need to choose numbers r and so
that r+s=b and rs=c.

a. Trial and Error
Example: 𝒙𝟐 – 𝟗𝒙 + 𝟐𝟎
There is no common factor so we begin by setting up
the first terms as follows: ( )(𝑥 )
Next we get the signs right. Since the last term is
positive, the signs will be the same as the middle
term which is negative. (𝑥 − )(𝑥 − )

Finally we determine the last term by trial
and error. The last terms multiplied
together should equal 20 and when added
equal −9.
Therefore our last terms will be −4 and −5.
(𝑥 − 4)(𝑥 − 5)

b. Difference of Two Square
Example: Factor x2 - 9
Both x2 and 9 are perfect squares. Since
subtraction is occurring between these squares,
this expression is the Difference of Two Squares
Answer: (x + 3) (x - 3) or (x - 3) (x + 3)

c. Perfect Square Trinomial
Example: Factor x2 + 2x + 1
Notice that x2 + 2x + 1 = x2 + 2x + 12
Using x2 + 2x + 12, we see that... the first term is
x2 and the base is x; the last term is 12 and the base
is 1
Put the bases inside parentheses with a plus
between them (x + 1)
Raise everything to the second power (x + 1)2

LINEAR INEQUALITIES
SOLUTION
SET
SET BUILDER
NOTATION
INTERVAL
NOTATION

A linear inequality describes a region of the coordinate
plane that has a boundary line.
Every point in the region is a solution set.
a < b and b > a
For two real numbers a and b, we say that a is less
than b, and write a <b , if there is a positive real
number p so that a + p = b. the statement b > a,
read b is greater than a, means exactly the same as
a < b.

If you add a positive number to any number the
sum is larger than the original number.
When we write a ≤ b we mean a < b or a = b
and say a is than or equal to b. when we write a
≥ b we mean a > b or a = b and say a is greater
than or equal to b.

A solution set is the set of values
which satisfy a given inequality. It
means, each and every value in the
solution set will satisfy the inequality
and no other value will satisfy the
inequality.

Example:
Solve 2x + 3 ≤ 7, where x is a natural number.
Solution:
2x + 3 ≤ 7 Subtracting 3 from both the sides
2x ≤ 4 Dividing both sides by 2
x ≤ 2 Since x is a natural number
Solution set = {1, 2}.

Shorthand used to write sets, often sets with
an infinite number of elements.
Note: The set {x: x > 0} is read aloud, "the
set of all x such that x is greater than 0." It is
read aloud exactly the same way when the
colon: is replaced by the vertical line | as in
{x | x > 0}.

General
Form:
{formula for elements: restrictions} or
{formula for elements| restrictions}
Examples {x: x ≠ 3} the set of all real numbers
except 3
{x | x < 5} the set of all real numbers
less than 5
{x2 | x is a real
number}
the set of all real numbers
greater than or equal to 0
{2n + 1: n is an
integer}
The set of all odd integers
(e.g. ..., -3, -1, 1, 3, 5,...).

INTERVAL
NOTATION
INEQUALITY
NOTATION
LINE GRAPH TYPE
𝑎 , 𝑏 𝑎 ≤ 𝑥 ≤ 𝑏 [ ] Closed
[ 𝑎, 𝑏) 𝑎 ≤ 𝑥 < 𝑏 Half – open
( ]a, b 𝑎 < 𝑥 ≤ 𝑏 ( ] Half – open
𝑎, 𝑏 𝑎 < 𝑥 < 𝑏 ( ) Open
[ 𝑏, ∞) 𝑥 ≥ 𝑏 [ Closed*
𝑏, ∞ 𝑥 > 𝑏 ( Open
(−∞, ]𝑎 𝑥 ≤ 𝑎 Closed*
−∞, 𝑎 𝑥 < 𝑎 ) Open
x
x
x
x
x
x
] x
x

PROPERTIES OF INEQUALITIES
ADDITION
PROPERTY OF
INEQUALITY
MULTIPLICATION
PROPERTY OF
INEQUALITY
DIVISION
PROPERTY OF
INEQUALITY
SUBTRACTION
PROPERTY OF
INEQUALITY

ADDITION
PROPERTY OF
INEQUALITY

Using Addition to Solve Inequalities
The solutions of an inequality like x < 3
are easy to recognize. To solve some other
inequalities, you may need to find a
simpler equivalent inequality.
Equivalent inequalities have the same
set of solutions.

Consider the inequality -4 < 1. The number line
shows what happens when you add 2 to each
side of the inequality.
-4 < 1
-2 < 3
-4 + 2 < 1 + 2
Notice that addition does not change the relationship
between the numbers of the direction of the inequality
symbol.

Addition Property of Inequality
For all real numbers a, b and c, if a > b, then a
+ c > b + c.
Example:
5 > -1, so 5 + 2 > -1 + 2
For all real numbers a, b, and c, if a < b, then a
+ c < b + c.
Example:
-4 < 1, so -4 + 2 < 1 + 2

EXAMPLE:
Solve x – 3 < 5. Graph the solutions on a
number line.
x – 3 < 5
x – 3 + 3< 5 + 3 add 3 to each side
x < 8 combine like terms
The solutions are all numbers less than 8.
Notation: 𝑥 𝑥 < 8
SS: (-∞.8)

SUBTRACTION
PROPERTY OF
INEQUALITY

Using Subtraction to Solve Inequality
Just as you can add the same number to
each side of an inequality, you can subtract
the same number from each side. The
order, or direction, of the inequality is not
changed.

Subtraction Property of Inequality
For all real numbers a, b, and c, if a > b, then
a – c > b – c.
Example:
3 > -1, so 3 – 2 > -1 – 2
For all real numbers a, b, and c, if a < b, then
a – c < b – c.
Example:
-5 < 4, so -5 – 2 < 4 – 2

EXAMPLE:
x + 2 < -6
x+ 2 – 2< -6 – 2subtract 2 from each side
x < -8 combine like terms
Notation: 𝑥 𝑥 < −8
SS: (−∞, −8)

MULTIPLICACTION
PROPERTY OF
INEQUALITY

Solving Inequalities Using Multiplication
You can find equivalent inequality by
multiplying or dividing each side of an
inequality by the same number. If the number is
positive, the order remains the same. If the
number is negative, you reverse the order.

Multiplication Property of inequality
For all real numbers a and b, and for c > 0:
If a > b, then ac > bc. If a, b, then ac < bc
Example;
4 > -1, so 4(5) > -1 (5)
-6 < 3, so -6 (5) < 3 (5)
For all real numbers a and b, and f or c < 0:
If a > b, then ac <bc.
Example:
2 > -1, so 4 (-2) < -1 (-2)
-6 < 3, so -6(-2) > 3 (-2)

EXAMPLE:
Solve:
𝑥
2
< -1. Graph the solutions on a number line.
𝑥
2
< -1
2 (
𝑥
2
) <2(-1) multiply each side by 2
x<-2 simplify each side
The solutions are all numbers less than -2.
Notation: 𝑥 𝑥 < −2 SS :( -∞, -2)

DIVISION
PROPERTY OF
INEQUALITY

Division Property of Inequality
For all real numbers a and b, and for c >0:
If a>b, then
𝑎
𝑐
>
𝑏
𝑐
If a<b, then
𝑎
𝑐
>
𝑏
𝑐
.
Ex.: 6> -4, so
6
2
>
−4
−2
-2 < 8, so
−2
2
<
8
2
For all real numbers a and b, and for c<0.
If a>b, then
𝑎
𝑐
<
𝑏
𝑐
. If a< b, then
𝑎
𝑐
<
𝑏
𝑐
.
Ex.: 6>-4, so
6
−2
<
−4
−2
-2<8, so
−2
−2
>
8
−2

EXAMPLE:
4d ≤ -28
4𝑑
4
≤
−28
4
divide both side by 4
d≤ -7 simplify each side
The solutions are all less than or equal to -7.
Notation: 𝑥 𝑥 ≤ −7
SS: (-∞, −7

COMPOUND INEQUALITIES
SOLVING COMPOUND
INEQUALITIES JOINED BY “AND”
SOLVING COMPOUND
INEQUALITIES JOINED BY “OR”

SOLVING
COMPOUND
INEQALITIES
JOINED BY “AND”

SOLVING
COMPOUND
INEQALITIES
JOINED BY “OR”

Solving Compound Inequalities Joined By “Or”
A solution of a compound inequality joined by or is any
number that makes either inequality true.
EXAMPLE:
Solve 4x = 3 < -5 or -2x + 7 < 1. Graph the solutions.
1
4
(4x) <
1
4
(-8) −
1
2
(-2x) >−
1
2
(-6)
x < -2 or x >3
The solutions are all numbers that are less than -2 or are
greater than 3.
SS: (−∞, -2 ) µ (3,∞)

ABSOLUTE VALUE EQUATIONS
AND INEQUALITIES

Absolute Value Equations
Use the following property to solve equations
that involve absolute value.
|x| = c is equivalent to |x| =± c
This property says that to solve an absolute
value equation, we must solve two separate
equations. For example, the equation |x| = 5 is
equivalent to the two equations x= 5 and |x| = -5.

EXAMPLE
Solve the equation |2x-5| = 3
Solution: the equation |2x-5| = 3 is equivalent to
two equations:
2x-5 = 3 or 2x-5 = -3
2x=8 2x=2 transpose 5 to
the right side
then simplify
x= 4 x=1 divide by 2
The solutions are 1 and 4.

Absolute Value Inequalities
We use the following properties to solve inequalities
that involve absolute value.

These properties can be proved
using the definition of absolute
value. To prove the property
one, that the inequality |x| < c
says that the distance from x to
0 is less than c and based from
the figure in left side you can
see that this is true if and only if
x is between c and –c.

Example
In solving an Absolute Value inequality
Sole the inequality |x-5|<2
Solution 1: The inequality |x-5|<2 is equivalent to
-2<x - 5<2 property 1
3 < x < 7 add 5
The solution set is the open
interval (3,7)
Solution 2 Geometrically, the
solution set consists of all numbers
x whose distance from 5 is less
than 2 and according to the figure
on the left side the interval is (3,7)

Polynomial Inequalities
A polynomial inequality is a mathematical statement
that relates a polynomial expression as either less
than or greater than another.
Example 1 – Graph:
Step 1: Write the polynomial in the
correct form. The polynomial must
be written in descending order and
must be less than, greater than,
less than or equal to, or greater
than or equal to zero.

Step 2: Find the key or critical
values. To find the key/critical
values, set the equation equal
to zero and solve.
Step 3: Make a sign analysis
chart. To make a sign analysis
chart, use the key/critical
values found in Step 2 to divide
the number line into sections.

Step 4: Perform the sign
analysis. To do the sign analysis,
pick one number from each of
the sections created in Step 3
and plug that number into the
polynomial to determine the
sign of the resulting answer. In
this case, you can choose x = –3
which results in +7, x = 0 which
results in –8, and x = 5 which
results in +7.

Step 5: Use the sign analysis
chart to determine which
sections satisfy the inequality.
In this case, we have greater
than or equal to zero, so we
want all of the positive
sections.
Step 6: Use interval notation
to write the final answer.

FINDING THE DOMAIN OF THE
FUNCTION

If a function is defined by an equation and the domain is
not indicated, then we assume that the domain is not the
set of all real number replacements of the independent
variable that produce real value for the independent
variables.
The domain of a function is the set of numbers that can
go in to a given function. In other words, it is the set of x-
values that you can put in to any given equation. The set
of possible y-values is called the range.

EXAMPLE
Find the domain of f(x) =
15
x−3
Solution:
The fraction
15
x−3
represents a real number for
all replacements of x by real numbers except x =3,
since division by 0 is not defined. Thus, f(3) does not
exist and the domain of f is the set of all real numbers
except 3. We often indicate this by writing f(x) =
15
x−3
x is not equal to 3.

To evaluate a function, simply replace (substitute)
the function's variable with the indicated number
or expression.
Example:
f(x) = 2x+4 for x=5
Just replace the variable "x" with "5":
f(5) = 2(5) + 4 = 14
10 + 4 = 14
Answer: f(5) = 14

1. If , find .
The notation means that everywhere we see x in
the function, we are going to replace it with 2. This
gives . Our notation is
• Notice that it is not absolutely necessary to
enclose the “2” in parentheses, but if functions are
more complicated it is helpful to avoid confusion.

Each function is defined for all x in the domains
of both f and h.
A. Sum of f and h is defined as:
The domain of f + h consists of the numbers x
that are in the domain of f and in the domain of
h.

B. Difference of f and h is defined as:
The domain of f-h consists of the numbers x that
are in the domain of f and in the domain oh h.
C. Product of f and h is defined as:
The domain of f·h consists of the numbers x that
are in the domain of f and in the domain of h.

D. Quotient of f and h is defined as:
The domain of f/h consists of the numbers x
for which h(x) not equal to 0 that are in the
domain of f and in the domain of h.

Illustrative example:
Let f(x)=x+1 and h(x)=2x2
1) The sum = (f + h)(x) = f(x) + h(x) = 2x2 x+1
2) The difference = (f - h)(x) = f(x) - h(x)
=( x+1)- 2x2= -2x2 + x+1
3) The product = (f . h)(x) = f(x) . h(x)
=( x+1)( 2x2) = 2x3+2x2
4) The quotient = (f/h)(x) = f(x)/h(x)
= x+1/2x2; h(x)not equal to 0

If f and g are functions, then the composite function, or
composition, of g and f is defined by
The domain of the composition function f o g is the set
of all x such that
• x is in the domain of g and
• g(x) is in the domain of f.

Let f(x)=2x-1 and g(x)
A. Find (f o g)(2) Solution find g(2)
Since g(x)
g(2)=
4
2−1
=
4
1
=4
4
1x


4
1x


EXAMPLE

Find (g o f) (-3)
Solution:
(g o f) (-3)= g(f(-3))= g(-7)
= 4/(-7-1)= 4/(-8)
= -1/2

Definition: An inverse function is a function
that undoes the action of the other function.
Remember:
The inverse of a function may not
always be a function!
The original function must be a one-
to-one function to guarantee that its
inverse will also be a function.

a. Given function f, find the inverse relation. Is the
inverse relation also a function?
Answer:
Function f is a one-to-one function since the x and
y values are used only once. Since function f is a one-to-
one function, the inverse relation is also a function.
Therefore, the inverse function is:

b. Determine the inverse of this function. Is the inverse
also a function?
x 1 -2 -1 0 2 3 4 -3
f (x) 2 0 3 -1 1 -2 5 1
Answer:
Swap the x and y variables to create the inverse
relation. The inverse relation will be the set of
ordered pairs:
{(2,1), (0,-2), (3,-1), (-1,0), (1,2), (-2,3), (5,4),(1,-3)}

Since function f was not a one-to-one
function (the y value of 1 was used twice),
the inverse relation will NOT be a function
(because the x value of 1 now gets mapped
to two separate y values which is not
possible for functions).

Finding the Inverse of a Function
Example:
Let f be defined by f(x)=3x+1. Find the inverse function f−1(x).
f(x) = 3x + 1 Steps:
y = 3x + 1 Replace f(x) by y.
x – 1 = 3y Interchange x and y.
Solve for y.
𝑥−1
3
=f-1(x) Replace y with f-1(x).
Therefore, we found out that x=y/3−1/3, so we
can write the inverse function as f-1(x) =
𝑥−1
3

Check:
First, we apply f followed
by f−1.
f−1∘f)(x) =f−1(f(x))
=f−1(3x+1)
=(3x+1)/3−1/3
=x+1/3−1/3
=x
Second, we
apply f−1 followed by f.
(f∘f−1)(x)=f(f−1(x))
=f(x/3−1/3)
=3(x/3−1/3)+1
=x−1+1
=x
In both cases, applying both f and f−1 to x gave us
back x. Indeed, f−1(x)=x/3−1/3.

A linear equation is an equation with two
variables whose graph is a line. The graph of the
linear equation is a set of points in the
coordinate plane that all are solutions to the
equation. If all variables represent real numbers
one can graph the equation by plotting enough
points to recognize a pattern and then connect
the points to include all points.

If you want to graph a linear equation you
have to have at least two points, but it's
usually a good idea to use more than two
points. When choosing your points try to
include both positive and negative values as
well as zero.

Graphing Linear Equations Using the Values of
x and y
A. Example:
Graph y = x + 2
Steps in Graphing :
1. Begin by choosing a couple of values for x
e.g. -2, -1, 0, 1 and 2 and calculate the
corresponding y values.

2. Evaluate y = 5 – 3x for different values of
x, and create a table of corresponding x and
y values.
X Y = x + 2 Ordered pair
-2 -2 + 2 = 0 (-2, 0)
-1 -1 + 2 = 1 (-1, 1)
0 0 + 2 = 2 (0, 2)
1 1 + 2 = 3 (1, 3)
2 2 + 2 = 4 (2, 4)

3. Plot the resulting ordered pairs in the
Cartesian plane.

4. Connect the points by drawing a line through
them.

Graphing Linear Equations Using Intercepts
Just remember:
To find the X Intercept: Let y = 0
To find the Y Intercept: Let x = 0
Example:
Graph the equation y = 3x + 2 by using the
intercepts method.

Steps:
1. Substitute y= 0 and solve for x.

2. Substitute x= 0 and solve for y.

3. Plot the two points and draw the graph.

GRAPHING FUNCTIONS USING
SLOPE-INTERCEPT FORM

Slope of a line- measures of its steepness
Slope =
vertical change(rise)
horizontal change (run)
SLOPE-INTERCEPT FORM:
y=mx+b
Slope y-intercept

HOW TO GRAPH:
1.) Plot the y-intercept on the y-axis. This is
the point (0, b).
2.) Obtain the second point using the slope.
Write m as fraction and use rise over run,
starting at the point containing the y-intercept,
to plot this point.

3.) Use a straightedge to draw a line through the
two points. Draw the arrowheads at the ends of
the line to show that the line continue indefinitely
in both directions.
y=
𝟑
𝟓
x-2
m=
𝟑
𝟓
b=-4
(0,-2)

Graph:
move 3 units up (rise)
move 5 units right (run)

Example:
1.)f(x)=2x-5 Given equation
y=2x-5 Change f(x) to y
m=2 b=-5 (0,-5)
x=2y-5 Substitute y to x and vice versa
-2y=-x-5 Isolate y-term
-
1
2
(2y=-x-5) Divide both sides by
1
3
y=
−𝑥
2
+
5
2
𝐹−1
(x) (inverse function)
m=
−1
2
b=
5
2
(0,
5
2
)

How to Graph:
1. Graph the f(x) by plotting the y-intercept on
the y-axis and obtaining the second point
using the slope.
2. Use straightedge to draw a line through the
two points. Draw the arrowheads at the ends
of the line to show that the line continue s
indefinitely in both directions.

3.) Graph the inverse function or the f-1(x).
4.) Use straightedge to draw a line through the
two points. Draw the arrowheads at the ends of
the line to show that the line continues
indefinitely in both directions.
5.) From the origin, draw a line in which the two
lines are intersect.

Graph:
y=x
z
𝐹−1
(x)=
−𝑥
2
-
5
2
f(x)=2x-5

Submitted by:
Allado, Swedenia
Artizona, Anamae
Bolivar, Ma. Cris
Annabelle
Callocallo, April Mae
Gabilagon, Liezl
Lerado, Aira Grace
Matalubos, Lyra
Sorenio, Shiela Mae
Tabuga, Natalie
T0rda, Marjocel
Submitted to:
Prof. Danilo Parreño
ADVANCE ALGEBRA

Advance algebra

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