Possible Intersection Multiplicities of a curve and hyperplanes

Let $$k$$ be an algebraically closed field with positive characteristic $$p>0$$ and let $$1 be its powers. Let $$X\subset \mathbb{P}^4$$ be given by $$(1:t:t^{q_1}:t^{q_2}:t^{q_2+1}+t^{q_1+q_3})$$. Then $$X$$ has orders $$0,1,q_1,q_2,q_2+1$$ i.e. there exist hyperplanes that have this intersection multiplicity with the curve.

The fact that the orders $$0,1$$ are correct comes from the fact that every curve has such intersection multiplicities. My problem is that I am unable to prove the rest formally. My guess would be that the hyperplane $$X_2=0$$ has intersection multiplicity $$q_1$$ with $$X$$ at the point $$(1:0:0:0:0)$$, however, I can't come up with a proof.

I would greatly appreciate your help!

The intersection multiplicity of a curve $$C\subset\Bbb P^n$$ and a hyperplane $$H$$ at a point $$c\in C$$ is defined to be the length of $$\mathcal{O}_{C,c}/(h|_C)$$ as a module over the local ring $$\mathcal{O}_{C,c}$$, where $$h$$ is a local equation for $$H$$ and $$h|_C$$ is its restriction to $$C$$. In your case, taking $$c$$ to be the point corresponding to $$t=0$$, the local ring $$\mathcal{O}_{C,c}$$ is a DVR with uniformizer $$t$$, and the hyperplane $$V(x_i)$$ restricts to the $$i$$th coordinate of your embedding. This means you can just read off the length from the lowest exponent present in each coordinate:
• $$V(x_0)$$ gives $$\mathcal{O}_{C,c}/(1)$$ which has length 0.
• $$V(x_1)$$ gives $$\mathcal{O}_{C,c}/(t)$$ which has length 1.
• $$V(x_2)$$ gives $$\mathcal{O}_{C,c}/(t^{q_1})$$ which has length $$q_1$$.
• $$V(x_3)$$ gives $$\mathcal{O}_{C,c}/(t^{q_2})$$ which has length $$q_2$$.
• $$V(x_4)$$ gives $$\mathcal{O}_{C,c}/(t^{q_2+1}+t^{q_1+q_2})$$ which has length $$q_2+1$$.