functions sd

7/25/2019 Functions SD

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FUNCTIONS

Domain: x elementsRange: y elements

We say that A is mapped to b and write as m: A B

f: x→x2 f: x→x + 3 g: x→x2 + 5x + 6

To find g(-1) we substitute -1 for xDomain : { -1,1,-2,2,-3,3} Domain :{ 0,1,2,3} Hence : g (-1)=(-1)2+5(-1)-6=-10

Range: {1,4,9} Range : {3,4,5,6} Similarly : g(0)=-6 and g(2)=8



We can find the domain

and the range of a function

from the graph :

Examples:

Domain: -1 ≤ x ≤ 4 Domain: x > 1

Range: 2 ≤ y ≤ 8 Range: y > 2

Domain: x є R Domain: {-2, -1, 1, 2 }Range: y є R Range: { 2 , 4 }



QUADRATIC FUNCTIONS

f(x) = ax2 + bx + c

Quadratic functions whose

graphs intersect the x - axis

can be written in the form :

y = a (x-p) (x-q)

x-intercepts (p,0) and (q,0)and if the curve touches the

x - axis in a point x1

then y = a ( x - x1)2 where

x = x1 axis of symmetry

The quadratic function can be

written also in the form

y=a(x-h)2 + k with vertex (h,k)



Examples:

a.

Axis of symmetry

1 2x x bx or x=

2a 2

(we will use the second because

we want to find x2 )

1 2

2

2

2

x x

x 2

1 x1

2

2 1 x

x 3 (-3,0)

b. i) vertex is at y = 5 and x = - 1

So f (- 1) = 5

ii) Range ( -, 5] or y 5



a. x2 - 2x - 3 =

( x - 3 )( x +1)

b. f(x) = x2 - 2x - 3

f(x)=( x-3)(x +1) but f(x) = a ( x - x1) (x - x2)

Hence A ( -1, 0)

B ( 3,0)

c.

1 2x x 3 1 2

x 12 2 2

b 2or x 1

2a 2 1

d.

2

b bVertex( , f ( )2a 2a

c (1, f (1)) (1, 4)

where f (1) 1 2 1 3 4



a) The value of c can be found if

x = 0

Then y = a 0 + 4 0 + c

y = c so c is the y - intercept c = 6

b) y = ax2

+ 4x + 6

The value of a can be found if we

substitute ( -1,0) or ( 3,0)

0 = a ( -1)2

+ 4( -1) + 6

0 = a - 4 + 6

0 = a + 2 → a = - 2

c) y = 2x + 4x + 6

We know that

a = -2 x1 = - 1 x2 = 3

y = a ( x - x1) ( x - x2)

y = - 2 ( x + 1) ( x - 3)



Example :

Find the intercepts

and the vertex of

y = 2 ( x - 2) ( x + 4)

Intercepts

x = 0 y =2 (0 - 2) ( 0 +4 )

y = 2 ( - 2 ) 4

y = - 16 ( 0, -16)

y = 0 0 = 2 ( x - 2) ( x + 4)

x1 = 2 x2 = - 4(2,0) ( - 4 ,0)

If the equation is y = 3 ( x + 3)2

then we understand that the graph

touches the x - axis at only one point

y = 0 x = 0

0 = 3 (x + 3 )2 y = 3 3

2

x = - 3 ( - 3 , 0) y = 27 ( 0 , 27)

Axis of symmetry

x = - 3

Vertex

Axis of symmetry 1 2x x 2 41

2 2

f(-1) = 2 ( -1-2) (-1+4)

= 2 (-3) ( 3 )

= - 18

Hence vertex = ( -1,-18)



EXPONENTIAL FUNCTIONS

I. The graph of f( x) = ax

f(x) = ax

when x = 0 y = ao = 1

Assymptate : y = 0

II. The graph of f(x) = aλx

Note that the graph of y =

2-x is the same as the

graph of y = ( ½)x because

2-x = (½)x



III The graph of the function f(x) = kaλx

Consider the

functions

y = 2x

y = 4 ∙ 2x

y = -2 ∙ 2x

All these functions

are asymptotic to

the x - axis

IV. The graph of the function f(x) = aλχ + k

Consider the functions

y = 2x

y = 2x + 2

y = 2x - 1

For every value of x

the value of y = 4 ∙ 2χ

is 4 times the y-value

of y = 2

x

For every value of x

the value of y = -2 ∙ 2χ

is the negative of the

corresponding y- value

of y = 2∙ 2x

For every value of x the value of y

for y = 2x + 2 is 2 units greater than

the corresponding y - value of y = 2x

( asymptote)

And the value of y for y = 2x -1 is 1

unit less than the corresponding y -

value of y = 2x (asymptote y = - 1)



Examples:

a. i. v = 32000r 0

v = 32000

ii. v= 32000r t

v = 32000 r 1

27200 = 32000 r

r = 0.85

b. 32000 ∙0,85t = 8000

0.85t = 8000/32000

0.85t = 0,25

t = 8.53 3 sf ( GDC)



The graph below shows the curve y k (2 x) + c , where k and c are constants.

Find the values of c and k .

C = -10 because y = -10 is thehorizontal asymptote

when x = 0 y = - 5

- 5 = k ∙ 2o - 10

- 5 = k - 10

k = 5

Hence = 5 ∙ 2x

- 10

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