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What is the rule for this function table?

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3 Answers

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.

The first thing to remember in this table is that the domain is the set of xs and the range is the set of ys or f(x)s.  First you should check and see if there is a constant slope for this table.  For each 1 in the domain I see the range increases 6.  So the slope is constant and equals the range/domain or here 6 over 1.  We have a linear equation since we have a constant slope.  Now I have to find out what the range equals when the domain equals 0, because I want my equation in slope intercept form, (y = mx+B, the y intercept is B and the slope is m) so I see that at x =1, y = 2, so if I go down 1 to x = 0, I have to go down 6 for my y, since the slope is 6, so y = -4, because 2-6 = -4, so my y intercept is -4.  The y interecept is what y equals when x = 0.  So my equation is y = 6x-4 or f(x) = 6x-4, since y =f(x).

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)