How to distinguish $y^2=(x+a)(x-b)^2$ from $y^2=(x+a)(x-b)^4$ if a graph is given? From the following question,

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.

 A: 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.
A: 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.
A: 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.
.
