What happens when we replace x with x+1 in a quadratic function?

[okunyg]

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. :)

 

[Werg22]

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.

 

[okunyg]

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

 

[Werg22]

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

 

[VietDao29]

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 + ... + b_n) = a.b₁ + a.b₂ + ... + ab_n

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. :)

 

[okunyg]

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.

 

[BSMSMSTMSPHD]

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.