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7/25/2019 Functions SD
http://slidepdf.com/reader/full/functions-sd 1/11
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
7/25/2019 Functions SD
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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 }
7/25/2019 Functions SD
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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)
7/25/2019 Functions SD
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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
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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
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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)
7/25/2019 Functions SD
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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)
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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
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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)
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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)
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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