Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.

[ReVeL]

Hahahaha.

[Collin Li]

No, I'm saying you guys have probably learnt about it already (preaching to the converted).

[vce01]

Let , so now you are looking at a graph.

Notice that it is a straight line:



You have enough data to solve for and .

You'll get:

Now substitute back in when you're done.

ok im a bit confused,

when you get the

y = mz + c thing

you just work out the y-intercept and gradient in the normal way right?

so shouldnt it be

(69-45)/(64-0) = m = 3/8

and c = 45 = y-intercept

giving,

y = 3/8 z + 45. but you guys are getting something else by the looks of it...

im probably wrong and this should be easy but im really slow today

[Collin Li]

Hmm... we concluded this question was wrong, but note that the left point is actually (40, 45), so you should get , and

[vce01]

ohh right, shoot, i thought it was (0, 45) -_- 

thankss for that lol

[ilovesuck]

^ ah yes, its quite poor they didnt put breaks in the axis...