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From the following question,

enter image description here

I understood that the answer must be one of options c and f. The other options are wrong (I will not explain why).

Question

How can I decide the correct answer from the left options (c and f)?

Edit

I plotted both with Mathematica as follows. However I prefer to solve it analytically without plotting them first if possible.

enter image description here

Display Name
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    Think about the difference between the curves for $ \ y = x^2 \ $ and $ \ y = x^4 \ $ at $ \ x = 0 \ \ . $ –  Sep 27 '21 at 08:24
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    @boojum: you pretty sure meant: compare $y^2 = x^2$ and $y^2 = x^4$ – orangeskid Sep 27 '21 at 08:26
  • What about in the immediate vicinity of $ \ x = 0 \ ? $ What happens with larger powers of numbers $ \ -1 < x < 1 \ \ ? $ –  Sep 27 '21 at 08:27
  • @orangeskid I chose to deal with $ \ y \ $ because the curves are more familiar. But there is a distinct difference in the shape of the curves near the $ \ x-$intercept that will tell us whether the exponent on $ \ (x-b) \ $ is $ \ 2 \ $ or $ \ 4 \ \ . $ The scale of the graph is sufficiently small to distinguish between the two choices. –  Sep 27 '21 at 08:36
  • All answers have been upvoted. Thank you very much for your contributions! – Display Name Sep 27 '21 at 09:25

3 Answers3

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Locally around $(b, 0)$, the term $x+a$ is positive, so $$ y^2 = (x+a) (x-b)^4$$ can be written as the union of two curves \begin{align} y &= \sqrt{x+a} (x-b)^2 ,\\ y &= -\sqrt{x+a} (x-b)^2. \end{align} and thus the derivative at $y = b$ is zero. But from your picture this is not the case. Thus $(f)$ is wrong.

Arctic Char
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The curve is a Tschirnhausen cubic . As detailed in the "Other equations" section in the wiki link, its equation can be represented by the form $27ay^2=(a-x)(8a+x)^2$, which can be adjusted to fit with the form as described in (c), by putting $a=-\frac 1 {27}$. So (c) would be the correct answer.

harry
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The curve $$C_n: y^2 =(x+a)(x-b)^n \tag{1}$$ intersects the $x$-axis twice. The right side of (1) is not only a function of $x$ and of $n$, but also of $a$ and $b$. The exact values of $a$ and $b$ don't really matter here, what's important to know is that they are both positive and thus

  • The intersection of $C_n$ with the negative real axis is caused by $a$.

  • The intersection of $C_n$ with the positive real axis is caused by $b$.

To investigate $C$ at $x=b$, the important thing to notice is that the contribution of the factor $(x-a)$ is not important because it's just like an almost constant factor of magnitude $b-a>0$ for $x\approx b$. Hence the curve is basically $$C_n:y^2 = c^2(x-b)^n \tag{2}$$ with some positive constant$^1$ $c$. So the curve can also be written as $$C_n:y = \pm c \,|x-b|^{n/2} \tag{3}$$

Let's have a look at $C_2$: $$C_2:y = \pm c \,|x-b|$$ You know how $x\mapsto |x|$ looks around zero, therefore $C_2$ close to $(b,0)$ looks like two crossing straight lines where the cross has the two symmetry axes $y=0$ and $x=b$.

On the other hand, $$C_4:y = \pm c \,(x-b)^2$$ looks like two kissing parabolas that are touching the $x$-axis in such a way that they are tangents, and the symmetry axes are the same as above.

For $n>2$ it's still kissing curves, but these cases are clearly visually distinguished from two crossing lines.

$^1$Using $c^2$ is just more convenient than using $c$ directly because we are goint to take squareroots.

.

emacs drives me nuts
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