GENERAL MATHEMATICS
ARNEL JR. A. MACUTAY, LPT
What we are about to learn..
a.Define functions and piece – wise functions
b.Determine whether a relation is a function or
not;
c.Represent real – life situations using functions
including piece – wise functions.
FUNCTIONS
Give another real – life situations involving functions.
Functions
A function is a relation in which each element of the
domain is paired with exactly one element of the range.
Another way of saying it is that there is only one output
(y) with each input (x).
Illustration of Functions:
a. set of ordered pair
b. mapping or arrow diagram
C. graphing
Functions
A function is a relation in which each element of the domain is paired
with exactly one element of the range. Another way of saying it is that
there is only one output (y) with each input (x).
Illustration of Functions:
a. set of ordered pair
1.{(3,2), (4,0), (5,1), (2,3)}
2.{(1,2), (0,3), (1,6), (5,4)}
3.{(3,4), (3,0), (3,1), (3,3)}
4.{(4,2), (3,2), (6,2), (5,2)}
1.Function
2.Not Function
3.Not Function
4.Function
Functions
A function is a relation in which each element of the domain is paired
with exactly one element of the range. Another way of saying it is that
there is only one output (y) with each input (x).
Illustration of Functions:
a. set of ordered pair (x, y)
x – First element DOMAIN
Independent Variable
y – Second element
RANGE
Dependent Variable
Activity #1
Find the domain and the range of the ff. set of ordered pair.
Do this in a ¼ sheet of paper.
1.{(1,2), (3,4), (5,6), (7,8)}
2.{(1,28), (2,29), (3,30), (4,31)}
3.{(3,a), (4,b), (5,c), (6,d), (7,e)}
4.{(2,9), (3,9), (6,4)}
5.{(3,2), (3,3), (3,4), (3,6)}
ANSWER!
1.x = (1,3,5,7) , y = (2,4,6,8)
2.x = (1,2,3,4) , y = (28,29,30,31)
3. x = (3,4,5,6,7) , y = (a,b,c,d,e,)
4. x = (2,3,6) , y = (9,4)
5. x = (3) , y = (2,3,4,6)
Illustration of Functions:
b. Diagram
To determine if y is a function of x, solve for y in terms of x.
x² – y = 4
x² – 4= y
Evaluate:
(1)² – 4= y, (2)² - 4 = y, (3)² - 4 = y,
(4)² - 4 = y
y = -3 y = 0 y = 5
y = 12
x = (1,2,3,4) y = (-
Example 1.1.1
Determine if the given equation represents y as a function of x.
a. x² – y = 4
Let’s try this!
Let g(x) = x² – 5x + 2
Find the following.
a.) g(-1)
b.) g(s)
c.) g(x - 1)
Let’s try this!
Let g(x) = x² – 5x + 2
a. g(-1) = (-1)² -
= 1 + 5 + 2
= 8
b. g(s) = (s) ² – 5(s) + 2
=s² – 5s + 2
Let’s try this!
Let g(x) = x² – 5x + 2
c. g(x-1) = (x-1) ² -5(x-1) + 2
= x² – 2x + 1- 5x + 5 + 2
= x² – 7x + 8
ACTIVITY #2
Evaluate the function at each specified value of the
independent variable and simplify.
1. g(x) = 5x -2
a. g(0)
b. g(-2)
c. g(x+1)
2. f(t) = -2t + 3
a. f(0)
b. f (1)
c. f( t + 1)
Exercise #1
Evaluate the function at each specified value of the
independent variable and simplify.
1. g(x) = 10x -5
a. g(10)
b. g(-4)
c. g(x+3)
2. f(s) = -2s + 20
a. f(-15)
b. f (10)
c. f( s -5)
PEMDAS(PARENTHESIS, EXPONENT, MULTIPLICATION,
DIVISION, ADDITION, SUBTRACTION)
1. (0)³ + 10 =
2. 15(5)²-50 =
3. 2(10)+30(5)-100 =
4. 125+9(15)+2(5)² - 2 =
5. 1(10)³+199+40(10)² - 8 =
Illustration of Functions:
c. Graphing
VERTICAL LINE TEST (PENCIL TEST)
If any vertical line passes through more than
one point of the graph, then that relation is
not a function.
THANKYOU
FOR YOUR
ACTIVE
LISTENING!

GENERAL MATHEMATICS First lesson for Grade-11. pptx

  • 1.
  • 2.
    What we areabout to learn.. a.Define functions and piece – wise functions b.Determine whether a relation is a function or not; c.Represent real – life situations using functions including piece – wise functions.
  • 4.
    FUNCTIONS Give another real– life situations involving functions.
  • 5.
    Functions A function isa relation in which each element of the domain is paired with exactly one element of the range. Another way of saying it is that there is only one output (y) with each input (x). Illustration of Functions: a. set of ordered pair b. mapping or arrow diagram C. graphing
  • 6.
    Functions A function isa relation in which each element of the domain is paired with exactly one element of the range. Another way of saying it is that there is only one output (y) with each input (x). Illustration of Functions: a. set of ordered pair 1.{(3,2), (4,0), (5,1), (2,3)} 2.{(1,2), (0,3), (1,6), (5,4)} 3.{(3,4), (3,0), (3,1), (3,3)} 4.{(4,2), (3,2), (6,2), (5,2)} 1.Function 2.Not Function 3.Not Function 4.Function
  • 7.
    Functions A function isa relation in which each element of the domain is paired with exactly one element of the range. Another way of saying it is that there is only one output (y) with each input (x). Illustration of Functions: a. set of ordered pair (x, y) x – First element DOMAIN Independent Variable y – Second element RANGE Dependent Variable
  • 8.
    Activity #1 Find thedomain and the range of the ff. set of ordered pair. Do this in a ¼ sheet of paper. 1.{(1,2), (3,4), (5,6), (7,8)} 2.{(1,28), (2,29), (3,30), (4,31)} 3.{(3,a), (4,b), (5,c), (6,d), (7,e)} 4.{(2,9), (3,9), (6,4)} 5.{(3,2), (3,3), (3,4), (3,6)} ANSWER! 1.x = (1,3,5,7) , y = (2,4,6,8) 2.x = (1,2,3,4) , y = (28,29,30,31) 3. x = (3,4,5,6,7) , y = (a,b,c,d,e,) 4. x = (2,3,6) , y = (9,4) 5. x = (3) , y = (2,3,4,6)
  • 9.
  • 10.
    To determine ify is a function of x, solve for y in terms of x. x² – y = 4 x² – 4= y Evaluate: (1)² – 4= y, (2)² - 4 = y, (3)² - 4 = y, (4)² - 4 = y y = -3 y = 0 y = 5 y = 12 x = (1,2,3,4) y = (- Example 1.1.1 Determine if the given equation represents y as a function of x. a. x² – y = 4
  • 11.
    Let’s try this! Letg(x) = x² – 5x + 2 Find the following. a.) g(-1) b.) g(s) c.) g(x - 1)
  • 12.
    Let’s try this! Letg(x) = x² – 5x + 2 a. g(-1) = (-1)² - = 1 + 5 + 2 = 8 b. g(s) = (s) ² – 5(s) + 2 =s² – 5s + 2
  • 13.
    Let’s try this! Letg(x) = x² – 5x + 2 c. g(x-1) = (x-1) ² -5(x-1) + 2 = x² – 2x + 1- 5x + 5 + 2 = x² – 7x + 8
  • 14.
    ACTIVITY #2 Evaluate thefunction at each specified value of the independent variable and simplify. 1. g(x) = 5x -2 a. g(0) b. g(-2) c. g(x+1) 2. f(t) = -2t + 3 a. f(0) b. f (1) c. f( t + 1)
  • 15.
    Exercise #1 Evaluate thefunction at each specified value of the independent variable and simplify. 1. g(x) = 10x -5 a. g(10) b. g(-4) c. g(x+3) 2. f(s) = -2s + 20 a. f(-15) b. f (10) c. f( s -5)
  • 16.
    PEMDAS(PARENTHESIS, EXPONENT, MULTIPLICATION, DIVISION,ADDITION, SUBTRACTION) 1. (0)³ + 10 = 2. 15(5)²-50 = 3. 2(10)+30(5)-100 = 4. 125+9(15)+2(5)² - 2 = 5. 1(10)³+199+40(10)² - 8 =
  • 17.
    Illustration of Functions: c.Graphing VERTICAL LINE TEST (PENCIL TEST) If any vertical line passes through more than one point of the graph, then that relation is not a function.
  • 19.