Hello, I need a little help here. I thought I understood this stuff but can't figure out this one concept. We're dealing with functions so here goes
f(x)=3x^2+2x-4
Find the value of f(x+1)
I thought it would work like this:
f(x+1)=3(x+1)^2+2(x+1)-4
=3(x^2+1)+2x+2-4
=3x^2+3+2x-2
=3x^2+2x+1
The answer is actually f(x+1)= 3x^2+8x+1
Can someone please explain this to me and tell me what i'm doing wrong?
2006-08-22 07:36:40 · 11 answers · asked by David T 1 in Science & Mathematics ➔ Mathematics
you only forget the (x+1)^2 is x^2+2x+1 and then you will multiply it by 3 so, you will find more 6x .
f(x+1)=3(x+1)^2+2(x+1)-4
f(x+1)=3(x^2+2x+1)+2x+2-4
f(x+1)=3x^2+8x+1
2006-08-22 07:45:44 · answer #1 · answered by emerzagor 2 · 1⤊ 0⤋
You had the first idea right:
f(x+1)=3(x+1)^2+2(x+1)-4
The problem was in the next step. You wrote:
3(x^2+1)+2x+2-4
But you have to square (x + 1)^2. It's not (x^2 + 1)
Its: (x + 1)*(x + 1) = x(x + 1) + 1(x + 1) = x^2 + x + x + 1
=x^2 + 2x + 1
f(x+1)=3(x^2 + 2x + 1) + (2x + 2) - 4
=3x^2 + 6x + 3 + 2x + 2 - 4
=3x^2 + 8x + 1
2006-08-22 16:51:55 · answer #2 · answered by Anonymous · 0⤊ 0⤋
f(x+1) = 3(x+1)^2+2(x+1)-4
= 3(x^2+2x+1) + 2x+2-4 <= you didn't square x+1 just x
= 3x^2 + 6x+2x + 3 +2-4
= 3x^2 + 8x +1
2006-08-22 07:45:29 · answer #3 · answered by TC 3 · 1⤊ 0⤋
I think I see where you went wrong with getting the "8x" part. If you write out every single step in full, you will see where you went wrong (see step 2).
f(x+1)=3(x+1)^2+2(x+1)-4
=3(x+1)(x+1)+2(x+1)-4
=3(x^2+x+x+1)+2(x+1)-4
=3x^2+3x+3x+3+2x+2-4
=3x^2+8x+1
2006-08-22 07:52:34 · answer #4 · answered by swhlye 2 · 1⤊ 0⤋
f(x+1) = 3 (x + 1)^2 + 2 (x +1) - 4
= 3 (x^2 + 2x + 1) + 2 (x + 1) - 4
= 3x^2 + 6x + 3 + 2x + 2 - 4
= 3x^2 + (6x + 2x) + (3 + 2 - 4)
= 3x^2 + 8x +1
you forgot to solve (x + 1)^2 before you distributed it to 3
2006-08-22 07:52:17 · answer #5 · answered by eyna 2 · 1⤊ 0⤋
f(x+1)=3(x+1)^2+2(x+1)-4 this line is ok
=3(x^2+1)+2x+2-4 ---> mistake 3(x+1)^2 = 3(x^2+2x+1) you miss 6x here so add it to your result
=3x^2+6x+3+2x+2-4
=3x^2+8x+1
2006-08-22 07:47:33 · answer #6 · answered by camedamdan 2 · 1⤊ 0⤋
here is the answer
f(x+1)=3(x+1)^2+2(x+1)-4
=3(x^2+2x+1)+2x+2-4
=3x^2+6x+3+2x+2-4
=3x^2+8x+1
give me 10p
2006-08-22 07:54:11 · answer #7 · answered by Anonymous · 0⤊ 0⤋
You got the right idea, but you just made a mistake with the (x+1)^2 term.
It should be:
(x+1)^2 = (x+1)*(x+1) = (x^2+2x+1)
2006-08-22 07:49:00 · answer #8 · answered by RobK 2 · 1⤊ 0⤋
correct assumption, wrong work
3(x + 1)^2 + 2x + 2 - 4
3((x + 1)(x + 1)) + 2x - 2
3(x^2 + 2x + 1) + 2x - 2
3x^2 + 6x + 3 + 2x - 2
3x^2 + 8x + 1
The (x + 1) aren't being squared individiually, the whole thing is being squared.
2006-08-22 11:53:28 · answer #9 · answered by Sherman81 6 · 0⤊ 0⤋
f(x+1)=3(x+1)^2+2(x+1)-4
f(x+1)= 3(x^2 + 2x + 1) +2x +2 -4
f(x+1)= 3x^2 + 8x +1
You didn't square x+1
2006-08-22 07:47:22 · answer #10 · answered by Jabberwock 5 · 1⤊ 0⤋