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Below, there is a plot of function: $a(x+d)^2 +b|x+d| +c$. What can you learn about coefficients $a,b,c,d$?

enter image description here

I know the answer: $a>0, b<0, c>0, d<0$ but I want to understand why is it so.

$a>0$ because (1) it is even-order polynomial and (2) its “tails” go up. Btw, there is nice chart summarizing that:

enter image description here

Could you help me with interpretation of other coefficients?

Igrek
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1 Answers1

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When $x$ is big positive or negative, our function is big positive. In the long run, our function is dominated by the $a(x+d)^2$ term, so $a$ is positive.

There is symmetry about $x=-d$. But there is symmetry about a positive $x$, so $-d$ is positive, and therefore $d$ is negative.

At $x=-d$, almost everything dies, we are left with $c$. But our curve is above the $x$-axis at $x=-d$ (the sharp point), so $c$ is positive.

At $x=-d$, the derivative of the $(x+d)^2$ part is $0$. So the slope of the curve near $x=-d$ is due to the $b|x+d|$ part. There, the curve looks kind of like $-|x+d|$. To see that, recall what $y=|x|$ looks like. So $b$ is negative.

Remark: The function is not a polynomial function, no curve $y=P(x)$ where $P(x)$ is a polynomial has a sharp bend (point of non-differentiability) like our curve.

André Nicolas
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  • I don't understand how do you know that sharp point is at $(x=-d)$? – Igrek Jun 30 '13 at 07:14
  • For example, for $f=(x-2)^2$ I see that its "bottom" is at x=2, but here we have part with modulus $b|x+d|$ – Igrek Jun 30 '13 at 07:16
  • We have symmetry about $x=-d$, because at $x=-d+\pm w$, the $a)x-d)^2$ part will give $aw^2$, the $b|x+d|$ part will give $b|w|$, neither cares about the sign of $w$. – André Nicolas Jun 30 '13 at 07:47
  • So the algebra says symmetry about $x=-d$. The geometry says symmetry about the sharp peak point. So the sharp peak point must be at $x=-d$. – André Nicolas Jun 30 '13 at 07:49
  • You are welcome. Note rhat symmetry, directly or indirectly, supplied a lot of information. The sharp point also helped a lot, because of the characteristic $|x|$-like behaviour, except upside down. – André Nicolas Jun 30 '13 at 08:02