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3 years ago

If f(x)= 2(3x²-2x+2). Find:A) f(x+1)B) f(3)C) f(y-2)

1 answer:

8 0

A) f(x+1) = 6x^2 + 8x + 6
1) substitute the value of x as (x+1)
2) 2(3(x+1)^2 -2(x+1) +2)
3) 6(x^2 + 2x + 1)
4) 6x^2 + 8x + 6

B) f(3) = 46
1) simplify the expression first
2) 6x^2 -4x + 4
3) Now just substitute 3: 6(3)^2 -4(3) + 4
4) 6•9 - 12 + 4 = 46

I hope this helps :)

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Write the frost ten terms of a sequence whose first term is -10 and whose common difference is -2

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Answer:

$-10,-12,-14,-16,-18,-20,-22,-24,-26,-28,...$

Step-by-step explanation:

we know that

In an Arithmetic Sequence the difference between one term and the next is a constant, called the common difference

The formula for an Arithmetic Sequence is equal to

$a_n=a_1+(n-1)d$

where

d is the common difference

n is the number of terms

a_1 is the first term of the sequence

In this problem we have

$a_1=-10\\d=-2$

substitute

$a_n=-10+(n-1)(-2)$

$a_n=-10-2n+2$

$a_n=-2n-8$

Find the first ten terms

$a_1=-10$

For n=2 ----> $a_2=-2(2)-8=-12$

For n=3 ----> $a_3=-2(3)-8=-14$

For n=4 ----> $a_4=-2(4)-8=-16$

For n=5 ----> $a_5=-2(5)-8=-18$

For n=6 ----> $a_6=-2(6)-8=-20$

For n=7 ----> $a_7=-2(7)-8=-22$

For n=8 ----> $a_8=-2(8)-8=-24$

For n=9 ----> $a_9=-2(9)-8=-26$

For n=10 ----> $a_1_0=-2(10)-8=-28$

The sequence is

$-10,-12,-14,-16,-18,-20,-22,-24,-26,-28,...$

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