Convert to class declaration quick fix misses superclass

<!-- 🚨 STOP 🚨 𝗦𝗧𝗢𝗣 🚨 𝑺𝑻𝑶𝑷 🚨 Half of all issues filed here are duplicates, answered in the FAQ, or not appropriate for the bug tracker. Even if you think you've found a *bug*, please read the FAQ first, especially the Common "Bugs" That Aren't Bugs section! Please help us by doing the following steps before logging an issue: * Search: https://github.com/Microsoft/TypeScript/search?type=Issues * Read the FAQ: https://github.com/Microsoft/TypeScript/wiki/FAQ Please fill in the *entire* template below. --> <!-- Please try to reproduce the issue with `typescript@next`. It may have already been fixed. --> **TypeScript Version:** 3.4.3 <!-- Search terms you tried before logging this (so others can find this issue more easily) --> **Search Terms:** - class declaration quick fix - 80002 **Code** This code is a mix of ES2015- and function-style classes: ```ts class A { constructor() { this.a = 'a'; } doAThing() { console.log(this.a); } } /** * @constructor * @extends {A} */ function B() { this.b = 'b'; } B.prototype = new A(); B.prototype.doBThing = function() { console.log('b: ', b); }; ``` The TS language services offer a quickfix to convert B into an ES2015 class. Great! **Expected behavior:** I'd hope that one of (or both of) the `@extends` and `B.prototype = new A();` would result in `B extends A`: ```ts class B extends A { constructor() { super(); this.b = 'b'; } doBThing() { console.log('b: ', b); } } ``` **Actual behavior:** The quick fix produces this code: ```ts /** * @constructor * @extends {A} */ class B { constructor() { this.b = 'b'; } doBThing() { console.log('b: ', b); } } B.prototype = new A(); ``` TypeScript flags an error on the last line: (property) B.prototype: B Type 'A' is missing the following properties from type 'B': b, doBThingts(2739) **Playground Link:** <!-- A link to a TypeScript Playground "Share" link which demonstrates this behavior --> no quickfixes on the…