You can use a derived version of the HKR theorem, i.e. $HH(A) = \text{Sym}_A^\bullet(\mathbb{L}_A[1])$ where $\mathbb{L}_A$ is the cotangent complex (over $k$). I'm not sure about a reference though.

Since $A$ is quasi-smooth, its cotangent complex will be in degrees $[-1, 0]$, and one can check using the fact that $\text{Spec}(A) = \{0\} \times_{\mathbb{A}^1} \{0\}$ that $\mathbb{L}_A \simeq \mathcal{N}^\vee_{\{0\}/\mathbb{A}^1}|_A \simeq A[1]$. Then, $HH(A) \simeq \text{Sym}_A^\bullet(A[2])$. You can do something similar for Hochschild cohomology using the tangent complex.

Now, the Connes $B$-operator is given by the de Rham differential. The module $\mathbb{L}_A$ is generated by $dx$, and the de Rham differential takes $x$ to $dx$. You can now calculate the various cyclic homologies, and in particular $HP(A) \simeq k(u)$ where $u \in H^2(BS^1; k)$ is the Chern class.

You can use a derived version of the HKR theorem, i.e. $HH(A) = \text{Sym}_A^\bullet(\mathbb{L}_A[1])$ where $\mathbb{L}_A$ is the cotangent complex (over $k$). I'm not sure about a reference though.

Since $A$ is quasi-smooth, its cotangent complex will be in degrees $[-1, 0]$, and one can check using the fact that $\text{Spec}(A) = \{0\} \times_{\mathbb{A}^1} \{0\}$ that $\mathbb{L}_A \simeq \mathcal{N}^\vee_{\{0\}/\mathbb{A}^1}|_A \simeq A[1]$. Then, $HH(A) \simeq \text{Sym}_A^\bullet(A[2])$ (note that this is in even degree, so as a chain complex it is $\bigoplus_{n \geq 0} A[2n]$). You can do something similar for Hochschild cohomology using the tangent complex.

Now, the Connes $B$-operator is given by the de Rham differential. The module $\mathbb{L}_A$ is generated by $dx$, and the de Rham differential takes $x$ to $dx$. You can now calculate the various cyclic homologies, and in particular $HP(A) \simeq k(u)$ where $u \in H^2(BS^1; k)$ is the Chern class (note that since $dx$ is in degree $-2$, it behaves like a symmetric variable not an exterior one).

You can use a derived version of the HKR theorem, i.e. $HH(A) = \text{Sym}_A^\bullet(\mathbb{L}_A[1])$ where $\mathbb{L}_A$ is the cotangent complex (over $k$). I'm not sure about a reference though.

Since $A$ is quasi-smooth, its cotangent complex will be in degrees $[-1, 0]$, and one can check using the fact that $\text{Spec}(A) = \{0\} \times_{\mathbb{A}^1} \{0\}$ that $\mathbb{L}_A \simeq \mathcal{N}^\vee_{\{0\}/\mathbb{A}^1}|_A \simeq A[1]$. Then, $HH(A) \simeq \text{Sym}_A^\bullet(A[2])$. You can do something similar for Hochschild cohomology using the tangent complex.

Now, the Connes $B$-operator is given by the de Rham differential. The module $\mathbb{L}_A$ is generated by $dx$, and the de Rham differential takes $x$ to $dx$. You can now calculate the various cyclic homologies, and in particular $HP(A) \simeq k(u)$.

You can use a derived version of the HKR theorem, i.e. $HH(A) = \text{Sym}_A^\bullet(\mathbb{L}_A[1])$ where $\mathbb{L}_A$ is the cotangent complex (over $k$). I'm not sure about a reference though.

Since $A$ is quasi-smooth, its cotangent complex will be in degrees $[-1, 0]$, and one can check using the fact that $\text{Spec}(A) = \{0\} \times_{\mathbb{A}^1} \{0\}$ that $\mathbb{L}_A \simeq \mathcal{N}^\vee_{\{0\}/\mathbb{A}^1}|_A \simeq A[1]$. Then, $HH(A) \simeq \text{Sym}_A^\bullet(A[2])$. You can do something similar for Hochschild cohomology using the tangent complex.

Now, the Connes $B$-operator is given by the de Rham differential. The module $\mathbb{L}_A$ is generated by $dx$, and the de Rham differential takes $x$ to $dx$. You can now calculate the various cyclic homologies, and in particular $HP(A) \simeq k(u)$ where $u \in H^2(BS^1; k)$ is the Chern class.