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Simple math help, need ASAP.

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asdfdsa
Posts: 1268 - Karma: 13
 describe how you determine the equation of a function of the form y=b^x, where b>0 if you are given its graph.
Nov 2 2009 11:10PM
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shragler
Posts: 572 - Karma: 32

ask allen. hes front right

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Nov 2 2009 11:12PM
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Jamestown
Posts: 4002 - Karma: 51
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Nov 2 2009 11:13PM
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pommes_vertes.
Posts: 1260 - Karma: 10
It's just a line with slope b that passes through the origin.  If I'm understood what you were asking, that is.
Nov 2 2009 11:14PM
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p-fo
Posts: 9372 - Karma: 6,155
y=mx+b? thats slope intercept form. i dont know what this carrot shit is. maybe its similar?
Nov 2 2009 11:15PM
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pommes_vertes.
Posts: 1260 - Karma: 10
Ah, fuck, I misread the question. 

y=b^x is an exponential graph.  Without a constant term, it'll pass through the point (0,1) and have a horizontal asymptote at y=0.  Also, the slope of the graph at any point is equal to the y-value, so given a graph, you should be able to figure out your b-value from that.
Nov 2 2009 11:21PM
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kingsskier
Posts: 6212 - Karma: 2,100
if b=1 it is a horizontal line passing through (0,1). Otherwise it has a y intercept at (0,1) and increases exponentially to the right of (0,1) and decreases to the asymptote on the horizontal line 0 to the left of the point. Ill see if I can get a picture for you.
Nov 2 2009 11:22PM
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kingsskier
Posts: 6212 - Karma: 2,100
this is if b is a constant term... If be is a variable, its completely different
Nov 2 2009 11:24PM
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blackoutbill
Posts: 2918 - Karma: 282
carrots=exponents

so y=b[to the]x is the same as y=b^x and the same as y=b[superscript]x

Nov 2 2009 11:24PM
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pommes_vertes.
Posts: 1260 - Karma: 10
That's pretty much it.

Essentially, it'll be in the form of y=b^x + c,  with c being the y-intercept.  Just look at the graph, find the y-intercept for your c-value, and then a couple other points (probably x=1, x=-1, x=2, x=-2, etc..), and use them to find your b-value.

For example, if you have points (0,4), (1,5), (2,13), (3,31) - you can see that your y-intercept, therefore your c-value, is 4.  From there, you know it's in the form of y=b^x + 4, and can find the b-value by looking at the other points --> b^1 + 4 = 5  --> b^2 + 4 = 13 --> b^3 + 4 = 31...

From there, you should be able to see that, in this case, b=3, and there you go.  y=3^x + 4

I hope that made sense.
Nov 2 2009 11:28PM
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kingsskier
Posts: 6212 - Karma: 2,100
he is right above with the +c. But I assume you dont need to worry about it because you didnt include it.
Here is a graph of z=2^x... all other graphs will look similar except x=1.
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Nov 2 2009 11:37PM
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kingsskier
Posts: 6212 - Karma: 2,100
sorry y=2^x not z even though it would work the same way as a variable
Nov 2 2009 11:39PM
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*mKas*
Posts: 52 - Karma: 11
He asked how to determine the function, not what the graph looked like, but the graphs are good anyway to help.

If you are determining a function of the form y=b^x, then you are determining the value of b based on the graph.  This is very easy.  When x=1, y=b.  So read the value of the graph at x=1 and that is the value for b.
Nov 3 2009 12:56AM
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[~_~]
Posts: 20239 - Karma: 1,273
holy shit
Nov 3 2009 2:56AM
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asdfdsa
Posts: 1268 - Karma: 13
Can you explain this any further? I'm really not even sure how that works. I know youve pretty much simplified it, but I dunno what you mean when you say y=b.. like i need to know how to get the transformations, and the rest of the shit
Nov 4 2009 9:29PM
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squibble
Posts: 4569 - Karma: 104
fuck, i just wrote some help and it didnt work
Nov 4 2009 10:03PM
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squibble
Posts: 4569 - Karma: 104
scan your graph and post it,

transformations are easy. in the form y=ab^x + c. c is the vertical displacement, it can be up or down (ex, if c= -2 shift your graph down 2 units.)

a is whether the graph is reflected in the x axis or stretched/compressed vertically. (if its 1 the graph doesnt change, if its -3 then reflect the graph in the x axis and stretch it by a factor of 3)

b determines whether the graph will be exponential growth, or exponential decay. google it. (its never negative)

if x has a negative sign in front of it then the graph is reflected in the y axis. (ex. -x + 3 will be a (in order) a shift to the left by 3 and THEN a reflection in the y axis.)

so if your equation is y= -3(4)^x-3 +6 shift right 3, reflect in the x, then shift up 6. you find the values of c and whats around x by looking at the graph.

to find a and b use two points on the graph, plug them in the equation y=ab^x + c, you will now have 2 equations (1 and 2 ) in equation 1 solve for a, then plug that into equation 2 to solve for b. put it all together and youll have your finished equation

that probably doesnt make sense
Nov 4 2009 10:16PM
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jamodeo
Posts: 3040 - Karma: 663
this is a simple math problem holy shit i havent been paying attention the last 2 years of high school
Nov 4 2009 10:26PM
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Ski4frnt4445
Posts: 420 - Karma: 132

what do you mean by transformations? like of the graph? because the transformations are y=t+- b^a(x+-h)

the t moves the graph up and down. the h moves the graph left and right. - numbers move it right and + numbers move it left. and a makes it grow and shrink

Nov 4 2009 11:18PM
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