About fibers of an elliptic fibration. Consider the following pencil of cubics:
$\lambda C_1+ \mu C_2$ where $C_1=y^2z$ and $C_2=x(x^2+2xz+z^2)$ and the elliptic fibration  $\tilde X \rightarrow \mathbb P^1$   induced by the blow-up of  $\mathbb P^2$ respect to the nine base points (counting multiplicity) of the pencil.
We have only two singular fibers corresponding to $C_1$ and $C_2$. 
Why these fibers are of type $I^*_0$?
 A: Why do you expect to get a relatively minimal elliptic fibration with singular fibres of type $I^*_0$?
In my opinion, blowing up base-points and eliminating fixed components produces a fibration $X \longrightarrow \mathbb P^1$ with $(-1)$-components in the two singular fibres. Blowing down these $(-1)$-curves produces the relatively minimal rational fibration $\Sigma_1 \longrightarrow \mathbb P^1$ of the first Hirzebruch surface. 

Each of the two cubic divisors $C$ and $D$ decomposes into a double line and a single line:
$$C:=V(y^2z) = V(z) \cup V(y^2),$$
$$D:=V((x^2+2xz+z^2)x) = V(x) \cup V((x+z)^2).$$
The resulting pencil of cubics $({\Gamma^1}_t)_{t = (t_0:t_1) \in \mathbb P^1}$ in $\mathbb P^2$ has $9 = 3 * 3$ basepoints with multiplicity
$$p_1=3*(0:1:0), p_2=4*(1:0:-1), p_3=2*(0:0:1),$$
see Fig. 1. The reduction of each irreducible component of $C$ and $D$ is isomorphic to $\mathbb P^1$. In the figures each component is annotated with its self-intersection number.
The blow-up at each base-point $\pi: X^2 \longrightarrow X^1 := \mathbb P^2$ contains the total transform $(\pi^*{\Gamma^1}_t)_t$ of the pencil $({\Gamma^1}_t)_t$ as a pencil with fixed component $E_1 + E_3$. Fig. 2 shows the strict transform 
$$({\Gamma^2}_t)_t:= (\pi^*{\Gamma^1}_t)_t - E_1 - E_3.$$
It has a base-point $p_4$ with multiplicity $2$ and a base-point $p_5$ with multiplicity $1$. 
An analogous blow-up and removing of fixed components gives the pencil from Fig. 3 with a single base-point $p_6$ with multiplicity 1. 
And a third blow-up gives the base-point free pencil $({\Gamma^4}_t)_t$ from Fig. 4. It is generated by the two divisors $C^4$ and $D^4$. Each has self-intersection $0$. Hence the resulting pencil defines a fibration $X:=X^4 \longrightarrow \mathbb P^1$ with distinguished singular fibres $C^4$ (blue) and $D^4$ (red). 
The surface $X$ results from $\mathbb P^2$ by blowing up 6 base-points. The surface is rational, its canonical divisor has self-intersection $\kappa_X^2 = 9-6 = 3.$
Apparently, the fibration $X \longrightarrow \mathbb P^1$ is not relatively minimal. The two singular fibres $D^4$ and $C^4$ blow down - by a total of 5 blow-downs - to a single fibre component $\mathbb P^1$, see Fig. 5 - Fig. 11. One obtains a rational fibration 
$$Y \longrightarrow \mathbb P^1$$
with $Y = \Sigma_1$ the first Hirzebruch surface. 


Note. The canonical divisor on $Y$ has self-intersection $\kappa_Y^2 = 3+5 = 8 = 9 -1$. From the Noether formula derives $c_2(Y) = \chi_{top}(Y) = 4$, in accordance with 
$$\chi_{top}(Y) = \chi_{top}(fibre) * \chi_{top}(base) = \chi_{top}(\mathbb P^1) * \chi_{top}(\mathbb P^1) = 2 * 2.$$  
