# Math Question (in Off-topic)

## Lord BobSeptember 9 2010 6:30 PM EDT

I took a math exam yesterday, and there was one question I couldn't answer for sure. Since I'm sure I aced the rest of the exam, it's been bugging me since then. Anybody here that can solve it for me?

Use the following points to write an equation in slope-intercept form:
(1, f(1))
(2, f(2))

## AdminQBGentlemanLoser[{END}]September 9 2010 6:56 PM EDT

I have a pure maths degree, and I've no idea! :P

## Lord BobSeptember 9 2010 7:15 PM EDT

So it's not just me then.

## Phoenix[The Forgehood]September 9 2010 7:49 PM EDT

slope is f(2)-f(1)
the equation is y=mx+b.
so you have y=(f(2)-f(1))x+b
plug in one point (1, f(1)) is easier
f(1)=(f(2)-f(1))(1)+b
b=2(f(1))-f(2)
therefore: f(x)=(f(2)-f(1))x+2(f(1))-f(2).
Or that's my understanding of it...

## El elSeptember 9 2010 8:01 PM EDT

Upon first looking I was going to say not enough information because you don't know what function you are using.

In my thinking it can be simplified to y=x.

Slope is going to be f(2)-f(1)/2-1 simplified to f(2)-f(1).

With the points (1, f(1)) and (2, f(2)) it goes with logical thinking that the y-intercept would be (0, f(0)). Therefore the answer would be y=(f(2)-f(1))x+f(0). Depending on the actual function it could be y=x but to be more general and possibly right I would stick with y=(f(2)-f(1))x+f(0).

I could be completely wrong, but that's my idea!

## El elSeptember 9 2010 8:09 PM EDT

@ above

"slope is f(2)-f(1)
the equation is y=mx+b.
so you have y=(f(2)-f(1))x+b
plug in one point (1, f(1)) is easier
f(1)=(f(2)-f(1))(1)+b
b=2(f(1))-f(2)
therefore: f(x)=(f(2)-f(1))x+2(f(1))-f(2).
Or that's my understanding of it... "

Plugging (1, f(1)) into y=(f(2)-f(1))x+b you'd get f(1)=(f(2)-f(1))(1)+b simplified to f(1)=(f(2)-f(1))+b to f(1)-b=(f(2)-f(1)) to -b=(f(2)-f(1))-f(1) to b=-((f(2)-f(1))-f(1)) since subtraction can't be flipped around the (f(2)-f(1)) must stay in ().

Would be y=(f(2)-f(1))x-((f(2)-f(1))-f(1)).

## AdminNightStrikeSeptember 9 2010 9:10 PM EDT

b is always f(0)

you can have:

y = (f(2) - f(1))x + f(0)

or you can solve for b using just the two given points:

y = (f(2) - f(1))x + 2*f(1) - f(2)

## SickoneSeptember 9 2010 9:43 PM EDT

y = (f(2) - f(1))x + 2*f(1) - f(2)

y = slope * x + intercept
where
slope = f(2)-f(1)
and
intercept = 2*f(1)-f(2)