hi good people!..

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!

juriemagic Feb 22, 2019

#1**0 **

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

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

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.

https://www.desmos.com/calculator/bqqgr7eyms

Here is the graph.

Melody Feb 22, 2019

#2**+1 **

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

#4**+1 **

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

juriemagic
Feb 22, 2019