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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Nomvalt on April 29, 2010, 08:10:53 am

Title: exponentials (easy)
Post by: Nomvalt on April 29, 2010, 08:10:53 am: For f(x) = 2^x, sketch the graph of each of the following, labelling asymptotes where appropriate:
a)y=f(x+1)
e)y=f(3x)
g)y=2f(x - 1)+1
h)y=f(x-2)

my questions:

for a), how is f(x+1) = 2*2^x?
for e), how is f(3x) = 8^x?
for g) how is 2f(x - 1)+1 = 2^x + 1?
for h) how is f(x-2) = 2^x/4?
Title: Re: exponentials (easy)
Post by: ZachCharge on April 29, 2010, 10:23:14 am: $f(x+1)=2^{x+1}$
$=2^x+2^1$
$=2*2^x$

$f(3x)=2^{3x}$
$=8x$
(First solve $2^3$ , and leave the x)

$2f(x-1)+1=2(2^{x-1})+1$
$=2(\frac{2^x}{2})+1$
$=2^x+1$

$f(x-2)=2^{x-2}$
$=\frac{2^x}{4}$

Remember, $2^{-1}=\frac{1}{2}$ and $2^{-2}=\frac{1}{4}$ (reciprocals).
Title: Re: exponentials (easy)
Post by: ZachCharge on April 29, 2010, 12:14:22 pm: Quote from: Nomvalt on April 29, 2010, 12:06:41 pm
How did you get from here:
Quote from: ZachCharge on April 29, 2010, 10:23:14 am
$2f(x-1)+1=2(2^{x-1})+1$
to here:
Quote from: ZachCharge on April 29, 2010, 10:23:14 am
$=2(\frac{2^x}{2})+1$

Where did the 2 in the denominator come from?

$2^{x-1}=2^x*2^{-1}$
As $2^{-1}=\frac{1}{2}$
$2^{x-1}=2^x*\frac{1}{2}$
$2^{x-1}=2^x*\frac{2^x}{2}$
Title: Re: exponentials (easy)
Post by: Nomvalt on April 29, 2010, 12:18:05 pm: Makes sense. :)
Dang, probably shouldn't have posted too quickly. Ah well...