**Domain: 1, 2, 3, 4. **

**Range: 2, 8, 14, 20**

**Domain: 1, 2, 3, 4. **

**Range: 2, 8, 14, 20**

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Kate,

The values give for x, the independent variable, are sequential 1, 2, 3, 4. So the values of y, the dependent variable, form a sequence for which we want the rule of formation. A technique that often works is to take differences:

original sequence: 2 8 14 20

differences: 6 6 6

So each term after the first is calculated by adding 6 to the previous term; start with 2.

Now let's find the formula for y in terms of x from this rule. The trick here is to notice that all except the first term are gotten by adding 6. This idea can be formulated as follows: each term is 6 times the number of steps away from the first term; that is, the term at x is 6*(x-1). Since the first term is 2, the formula for y is

y = 2 + 6*(x - 1)

There are other ways to solve this problem that make sense in different contexts. I don't know your context, so I don't know how to recommend you solve such problems.

Joseph M. | A Former Math Teacher who is Mr. Renaissance!A Former Math Teacher who is Mr. Renaiss...

When it wants a rule for a function you have to ask yourself

“what do I have to do to x to get those corresponding y values”

Each y (y1,y2,…) will represent a different attempt

x 1 2 3 4

y1 2 8 14 20

y2 2 4 6 8

y3 6 12 18 24

y4 2 8 14 20

y1 = the values given

This is a little bit a game of guessing. For instance my thought process went:

y2: does y = 2x?

2x works for getting when x=1 y=2, but not for any of the others because you get the values for y2 (above)

My y’s are off, but how far off? Do each y1 – y2:

2-2 = 0, this is perfect

8-4 = 4, so 4 off

14-6 = 8, so 8 off

20-8 = 12, so 12 off

Now I can see a pattern of the y’s increasing in increments of 4, but how much did I need to multiply that 4 by each time before adding it to the 2x value?

y3: deos y = 2x +4x?

By the same x value? No, that gives me a number that is too big

y4: does y = 2x + 4(x-1)?

Yes, this works for all values, so this can be the rule

**2x + 4(x-1)**

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