+1124 hi good people!.. 
I am given g(x)=a.bx+q
points on the graph are C(0;1) and E(−1;−13)
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!
+118713 1=a+qq=1−a
−13=ab−1+q−13=ab+qb≠0−13=ab+1−a3b(−13)=3b(ab+1−a)−b=3a+3b−3ab0=3a+4b−3aba and b both equal 0 or−3a(1−b)=4b3a(b−1)=4ba=4b3(b−1)b≠1
If b =1 then
g(x)=a+qg(x)=a+1−ag(x)=1
but then point G is not on the curve so b cannot equal 1
So we have
.q=1−aa=4b3(b−1)whereb≠1b≠0
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.
+1124 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 12,
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?
+1124 yes, I saw that, thank you so much for your time...I do apreciate!..as always!..
