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7/26/2019 13.2 Lesson - KEY
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13.2 Lesson.notebook
Name
Class
Date
13,2 Exponential Growth Functions
Essentia| Qu@stion: How is the graph of g(x) = ab- h + k where b > 1 related to the graph
of f(x) = b"?
[]
J
x
r=
Resource
A;t,S,A Determine the effects on the key attributes on the graphs of f(x) = b... where b is Lm:ker i
2, I 0 . . . . when f ix' is replaced by of(x), f {x) + (t, and f(x - c) t 'or specif ic posit ive and
negative real values of a0 c, and d. Also A2.2.A, A2.5,B, A2.5.D, A2.7.1
Graphing and Analyzing f(x) = 2"
and f(x) = 10x
A m ' t t expo t l e t t t id f ime t ion is a func t i on o f the to tn he re b i s a pos i ti ve cons tan i o the r than I and t le
exponent x s a variable. Notice that there is no single ent exponential function because each choice of the base b
determines a different parent function,
Complete the input-output tab le lk , r each of the parent exponent ia l l \mct ions be low.
l
i
o I
3 I } #5C 5
Graph the parent functions J-'(x) = 2' and p(x) = 10" by plotting points.
---- ......... 7-...i. I-". -/
-i --T , .
Modu,,, 1.3 713
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Ma rch 1 O, 2016
What i s the domain o f ea ch func t i on?
Don,ainoff(x)=2k{.x,_-c*>}
Domain ofp(x)= 10}
What is the range of each f imct ion?
What i s t he : -i n t e rcep t o f each func tion?
yAntercept of f(x) --- 2': (O,Y [ )
y intercept o f p(x ) = lOq (O, ' )
"What is the t rend of each funct ion?
In both ] ' (x ) = 2 and p(x) = 10, as the
w,l.o o
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,13.2 Lesson.notebook March 10, 2016
Pred ic t what the graph o f ea ch f imct ion w i ll looklike , and then ske tch the graph bas ed on the graph shown on your
calculator.
Q The grap h ofg(x) = (2") wi I1 be t i le graph of f (x)= 2 ' ver t ical ly g 'e) , ' : ". - :?(+:""t y. . a
factor of2 ,
I :hb raph o fg,(x) = 3(2') , , , il l be the graph off(x) = 2' vert ically 't J[.7[ : tt '. ([-bya
factor o _ ,
The graph ofqt(x) = 2(1 (V) wi l l be the graph ofp(x) = 10" ver t ical ly wl (=.r] rP- f f by a
factor o f . . 5)
2
The graph ofq2(x)= (10")will be the graph ofpx) = 10 vertically fOt%{:'2[v'l by a
factor of_]__,
...... i--4
i ii i
:i 1
I--L;.=I c
5
@ The graph ofgdx) = -(2' ) will be the graph off(x) = 2 reflected across the 77 % ;":-.-
and vertically by a factor ot',
The graph ofgz(x) = -5(2' ) will be the graph of fix) = 2* reflected across tile .. _7 ,-i.4.-.
and vert cally by a S
actor of .........
The graph ofqt{x) .... {o(1(.)) will be the graph ofp(x) = t0" reflected across the+ O./'5,
and vert ically ,5'J'g:+g L.d_.. by a factor of__.. .+.
--g(lO ) will be tile graph ofp(x) = 10 rellected across the
he graphofq,(x) = 1
and vertically, by a factor of .
I. i ::
4--[
{ ' _1 ..... ......
L_I i_: i_ i
Modulv t3
7 1$
L , s . . s o v 2
3
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The graph ofg,(x) = 2 +' will be the graph off(x) = 2' translated ..... .. .. unit to the _{t{ i . }_i. ..... ....
The graph ofg,(x) = 2 " 4 will be the graph off(x) = 2:' translated t [the (q te ?"
unts to .:
The graph ofq(x) 10' +2 will be the graph ofp(x) = 10' translated unit to the I l
The graph ofq2(x) = 10'-:' will be the graph ofp(x) = t0-' translated ....... units to the .......... Q .....
U -%
i
7-i T-T]
/5_J
yi Ul7--i
i 18:
i: 114-
i . . . . .
px)-:-;i
: : i :
I '
@ The graph of gl(x) =2 + 3 illbe thegraph olfi, x) = 2 translated J units
The graph of&(x) = 2' -- 5 ,,.,ill be the graph offix) = 2 translated .... :[;-, units
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Graphing Combined Transformations of f() = bx
Wher e b > 1
A given expone nt ia l f imct ion g(x) = a(b ' - ' ) + k wi th base b can be graphe d by reco gniz ing the d i f ferences betwe en
the given function and its parent t;anction, fx) = bL These differences define the parameters of the transformation,
where k represents the vertical translation, h is the horizontal translation, and a represents either the vertical stretch
or compression of the exponentkd function mad whether it is rellected across the x-a.ds,
-o ,. _
You can use the parameters in g(x) ---- a(F' -- h) + k to see what happens to two reference points during a
transformation, '1\vo points that arc easily visualized on the parent exponential fimction are (0, I) and (I, b),
m . t ransrorma.on, t l , po in t (0 , ) b, ,mes ( t , , , , + k) a.d 0 , b) beome, ( + h , , ,b + k) . The a. . rmptote y = 0 for
the parent fmaction becomes y = k
The graphs off(x) = 2' and p(x) = 10' m'e shown below with the reference points anti asymptotes labeled,
i l i i + i i
>-I-i-i--6- 7 .... ----I
(o,1)y.- Ox;
{r-4_--2 PL. 2 4
{+ 8
i
i2
i + + i
i i
/o,, +L ,
Example I State the domain and rmage of the given function, Then identif7 the new
values of the reference points and the asymptote. Use these values to graph
the function.
&
t
P
t2
g
I
g(x) = --3(2x-') 4" 1
ho,-,o,non =+,s b - +o
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( I , b) becomes ( l + h, ab + k) ,
(t +/,,,,b+k)=(t+2,-3(2t+l)
=(3,-6+1)
= (3,- 5)
The asympt61e becomes y = k.
y=k --* ),=1
Plo t the t r ans tb rmed po in ts and a s ympto te and d raw the cu rve .
( q(x) = 1.s(lo'-') - s
Examine q(x ) and i den t if y the pa ramete rs ,
a = /: so the functmn is stretched vertical_ty by a factor of 1.5.
h so the function is translated 3 unils to the right.
The asymptote becomes y = k,
y=k -- ),=-
Your Turn
7. c*) = 4(2+') - 6
Modu le 13
7 18
Plot the translbrmed pohts alld asymptote and draw the curve,
-{--+ ....... :- ......... X l
=a
g
9
3
(]
[
] A:s,ol 2
6
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8
q(x @L (lo. o + 3
L'-_( . ":-.4._: i :1"=2
: : ":- i " T ....
: L.2 J. ,
'=T "F4:
d-i
:J .... .:if:
Writing Equations for Combined Transformations
o f f (x ) = b Whe r e b > 1
c
J
E
o
Given the graph of an exponentia fimction, you can use your knowledge of the transtbrmation parameters to write
the function rule for the graph, Recall that the asymptote will give the value ofk and the x-coordinate of the first
re;renee point is h, Then let ), he they-coordinate ofthe first point and solve the equation y = a + k tbr a,
Finally,, use a h, and k to write the function in the form g(x) = a (b ') + k,
Example 2
Write the exponential function that wilt produce the given graph, using
the specified value of b. Verify that the second reference point is on the
graph of the function. Then state the domain and range of the function in
set notation.
Let b = 2,
The asymptote is y = 1, showing that k : 1.
The first referen:e point is (--I,-I). This shows that h = -land
tha a +k : -- " "
3"
Substitute k = 1 and solve for a.
a+k=-
3
I
a+:----
3
4
h = -
3
k-l
S ubs t it u te t hese va lues in t o g (m = a (b - ) + k t o f i nd g (x ) .
X(x) = a(b-') + k
3
i.
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t "2
er i fy tha g{-/ J = _5
= 4_(20 +
3
4
=/__ (2) + 1
3 "8
3 3
5
}.
Let b = 10,
The asymptote isy = [(5 , showing that k = 0 .
The first reference point is (=4, 4,4). This shows that h --- - t/and
that a + k = ],:/. Substitute for k and solve for a,
,, + k= "I,,/
,,+/9 = ,'/
.=/,:
h = L[
Substitute these values into q(x) = a(bx-') + k to find q(x) .
,,.,.=,,I'-"l k= "/,1o (,o":) + I'
i--- . =-:.....: A 7. f y.i.- i
,i i ' ' i,
" i-kSl-- :
Verify that q(--3) = --10,
=o7:Io (,o')+
=ito + t.,
The domain ofq(x) is
The range o f q (x ) i s
g
::r
3
s
r
Modu le 13
7 2 0
Liassoll 2
8
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Your Turn
%trite the exponential function that will produce the given graph, using the specified
value of b. VcrifX that the second reference point is on the graph of the function, Then
state the domain and range of the function in set notation,
1 "'> i " , i ,
" " , i Ii i I
,, (4,w
I -: i x
. J_:__2 .t 3.Z;4i
--;2. ,
JJ
>< I --"