<p>I'm looking at a chart right now so I'll just post it in this form:</p>
<p>n=1 f(n)=7
n=2 f(n)=13
n=3 f(n)=19
n=4 f(n)=P
n=5 f(n)=31
n=6 f(n)=37</p>
<p>a)21
b)23
c)25
d)27
e)29</p>
<p>This is an easy question so I sort of feel bad that I wouldn't know how to do it if it were on an actual test the algebraic way. I can see that it's 6n+1, but how do you get that? I got that by just looking at the #s, but if a linear function turns up as a hard question, I'll need to know the algebraic way. </p>
<p>What are you supposed to do, do the y1-y2/x1-x2 to get a slope and set that equal to 2 other values(with one of them being p)?</p>
<p>Just look at the pattern (change in y) of f(n), which is +6 to consecutive terms. 19+6=25</p>
<p>But you can't always just look for the pattern. How do you do it using the slope method? You would need to know the slope method if they used linear functions as a hard question because they wouldn't make a hard question solvable by looking for a pattern.</p>
<p>find slope from one point to another point. You find that m = 6</p>
<p>y = mx + b
7 = 6x + b, solve for b = 1
y = 6x + 1</p>
<p>Thanks! I'm just trying to cover all the hard concepts that I can because I want to get an 800. The whole idea is to aim for the highest, no point in aiming for a score like 700 because then your mentality will be to omit some hard questions. If you aim to get all the questions, then you have a better chance of getting them all. I've averaged 730 on two REAL SAT Math sections that I've taken(January and October 2007 of this year that my friends let me borrow). I feel that I can get a very good score on the real SAT that I'll be taking in January, but I'm trying to make sure that I don't get stumped by concepts that I'm not familiar with.</p>