# Something extremely easy

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

Thanks. :)

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.

Well, sure. But could you tell me if I was at least thinking correctly?

Yes you were. Look up for "distributive law" on google to get more insight.

VietDao29
Homework Helper
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

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. :)

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.

You might also try putting it in standard form: $$y=a(x-h)^{2}+k$$ where $$(h,k)$$ is the vertex of the parabola. Then, replacing $$x$$ by $$x+1$$ simply shifts the curve 1 unit to the left (ie, decreases $$h$$ by 1). The process is called completing the square.

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

$$5f(x) = 80000 + 250x + x^2$$

$$5f(x) = 80000 - 15625 + 15625 + 250x + x^2$$

$$5f(x) = 80000 - 15625 + (x + 125)^2$$

$$5f(x) = 64375 + (x + 125)^2$$

$$f(x) = 0.2(x + 125)^2 + 12875$$

So, the vertex is $$(-125, 12875)$$. Replacing $$x$$ by $$x+1$$, we get:

$$f(x + 1) = 0.2(x + 126)^2 + 12875$$

which has a vertex of $$(-126, 12875)$$, one unit to the left of the original.

The nice thing about this method is that you can replace $$x$$ with $$x + anything$$ and it's already simplified.

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