Fill in the table using this function rule y=3x+3

Table:

(x,y)

(-6,-15)

(-3,-6)

(0,3)

(3,12)

Step-by-step explanation:

If the function rule is y = 3x + 3, for each x value, we fill it into the formula.

(Tip: when I put x when solving I mean multiply)
1. -6

Fill in: y = (3) -6 + 3

Solve:

multiply: 3 x -6 = -18

add: -18 + 3 = - 15

Table way: (x,y) = (-6,-15)

2. -3

Fill in: y = (3) -3 + 3

Solve:

multiply: 3 x -3 = -9

add: -9 + 3 = -6

Table way: (x,y) = (-3,-6)

3. 0

Fill in: y = (3) 0 + 3

Solve:

multiply: 3 x 0 = 0

add: 0 + 3 = 3

Table way: (x,y) = (0,3)

4. 3

Fill in: y = (3) 3 + 3

Solve:

Multiply: 3 x 3 = 9

add: 9 + 3 = 12

Table way: (x,y) = (3,12)

yes

yes yes yes

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}