Something extremely easy

  • Thread starter okunyg
  • Start date
  • #1
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Too bad I've forgot how to do this though. I'll paste the problem:

f(x) = 16000 + 50x + 0.2x^2

What happens if x = x+1 (increase by one)?

I believe it's something like this:

50(x + 1) + 0.2(x+1)^2 ->
50x + 50 + 0.2(x^2 + 2x1 + 1^2) ->
50x + 50 + 0.2x^2 + 0.2x + 0.2 ->
(I'm not sure about the next step because of my bad algebra skills): Possibly ->
50.2x + 50.2 + 0.2x^2
or
50x + 50.2 + 0.4x^2

Could anybody please help?
Thanks. :)
 

Answers and Replies

  • #2
1,425
1
Not to be mean or anything, but it would be best if you found out by yourself... something like this is better not answered by others.
 
  • #3
17
0
Well, sure. But could you tell me if I was at least thinking correctly?
 
  • #4
1,425
1
Yes you were. Look up for "distributive law" on google to get more insight.
 
  • #5
VietDao29
Homework Helper
1,423
2
Too bad I've forgot how to do this though. I'll paste the problem:

f(x) = 16000 + 50x + 0.2x^2

What happens if x = x+1 (increase by one)?

I believe it's something like this:

50(x + 1) + 0.2(x+1)^2 ->
50x + 50 + 0.2(x^2 + 2x1 + 1^2) ->
50x + 50 + 0.2x^2 + 0.2x + 0.2 ->
(I'm not sure about the next step because of my bad algebra skills): Possibly ->
50.2x + 50.2 + 0.2x^2
or
50x + 50.2 + 0.4x^2

Could anybody please help?
Thanks. :)
The first 2 lines are correct. However, the 3rd is not.

Ok, I am pretty sure that you have covered the distributive property of multiplication over addition, right?
It goes like this:
a . (b1 + b2 + b3 + ... + bn) = a.b1 + a.b2 + ... + abn

Ok, in your problem, we have:
50 (x + 1) + 0.2 (x2 + 2x + 1)

Apply the above property, we have:
... = 50x + 50 + 0.2 x2 + 0.2 * 2 x + 0.2
= 50x + 50 + 0.2 x2 + 0.4 x + 0.2

I believe you can take it from here. :)
 
  • #6
17
0
I really, really need to brush up on my algebra.

50x+50 + 0.2x^2 + 0.4x + 0.2

I believe the correct answer was 50.2 + 0.4x. I can't look in my book since a friend borrowed it.

50.4x+50.2 + 0.2x^2 is my answer

I'm not able to solve it. Could you do me a favor and help me out a little? I'm not actually studying math at the moment, I'm just playing around, thus my knowledge is at a bad state.
 
  • #7
131
0
You might also try putting it in standard form: [tex] y=a(x-h)^{2}+k[/tex] where [tex] (h,k) [/tex] is the vertex of the parabola. Then, replacing [tex]x[/tex] by [tex]x+1[/tex] simply shifts the curve 1 unit to the left (ie, decreases [tex]h[/tex] by 1). The process is called completing the square.

[tex]f(x) = 16000 + 50x + 0.2x^2[/tex]

[tex]5f(x) = 80000 + 250x + x^2[/tex]

[tex]5f(x) = 80000 - 15625 + 15625 + 250x + x^2[/tex]

[tex]5f(x) = 80000 - 15625 + (x + 125)^2[/tex]

[tex]5f(x) = 64375 + (x + 125)^2[/tex]

[tex]f(x) = 0.2(x + 125)^2 + 12875[/tex]

So, the vertex is [tex](-125, 12875)[/tex]. Replacing [tex]x[/tex] by [tex]x+1[/tex], we get:

[tex]f(x + 1) = 0.2(x + 126)^2 + 12875[/tex]

which has a vertex of [tex](-126, 12875)[/tex], one unit to the left of the original.

The nice thing about this method is that you can replace [tex]x[/tex] with [tex]x + anything[/tex] and it's already simplified.
 
Last edited:

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