hi good people!.. smiley


I am given \(g(x)=a.{b^x}+q\)


points on the graph are \(C(0;1)\) and \(E(-1;{ -{1\over3}})\)


Determine a, b and q.


Okay, I know q is the y-intercept, so that is 1

so, now we plug in the values of any point to calculate b.

If the "b" was alone, I would be able to do it, but there is an "a" multiplied with the "b" as well....uhm, I'm slightly confused...can anyone please teach me the way?..


I do thank you all!

 Feb 22, 2019

\(1=a+q\\ q=1-a\)



\(\frac{-1}{3}=ab^{-1}+q\\ \frac{-1}{3}=\frac{a}{b}+q \qquad \qquad b\ne0 \\ \frac{-1}{3}=\frac{a}{b}+1-a\\ 3b(\frac{-1}{3})=3b(\frac{a}{b}+1-a)\\ -b=3a+3b-3ab\\ 0=3a+4b-3ab \qquad \text{a and b both equal 0 or}\\ -3a(1-b)=4b\\ 3a(b-1)=4b\\ a=\frac{4b}{3(b-1)} \qquad \qquad b\ne1\\ \)


If    b =1    then

\(g(x)=a+q\\ g(x)=a+1-a\\ g(x)=1 \)

but then point G is not on the curve so  b cannot equal 1


So we have


.\(q=1-a\\ a=\frac{4b}{3(b-1)}\\ where \;\; b\ne1 \;\;\;\; b\ne0\)


PLUS, I am not so sure that b can be negative........   

The graph would be very broken if b was negative, I am not sure how much that matters.

Maybe another mathematician would like to comment here. 


Anyway ,...There is no single solution for this.


Here is the graph.




Here is the graph.

 Feb 22, 2019

Hi Melody,


According to the handbook, "b" can never be negative. It can only be either a fraction between 0 and 1, or greater than 1. Okay, so this I accept. There is an example in the book on a similar sum, but it is not clear to me. By the way it looks, it appears the "a" is disregarded and "b" is calculated. I followed that aproach and found "b" to be \(1 \over2\),

I then substituted this also into the equation and found "a" to be -1. But I'm not sure if that was the right way to go about it?

juriemagic  Feb 22, 2019
edited by juriemagic  Feb 22, 2019

But my graph shows that a can vary as well.   frown



Melody  Feb 22, 2019

yes, I saw that, thank you so much for your time...I do apreciate!..as always!.. smiley

juriemagic  Feb 22, 2019

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