g | x | w | all
Bytes Lang Time Link
038Zsh241221T095826Zroblogic
087brainfuck241220T135333ZWeirdGly
050Python 3241205T205028ZFmbalbue
034AWK241206T184421Zxrs
028R241205T142306ZEonema
027R241205T133445Zpajonk
008APL+WIN241205T101702ZGraham
007Charcoal241205T105326ZNeil
046PowerShell241205T104828Zuser3141
00205AB1E241205T102119ZKevin Cr
042JavaScript Node.js241205T094207Zl4m2
005APL Dyalog 20.0241205T093817ZAdá
010Vyxal 3241205T094007ZThemooni

Zsh, 38 bytes

for m ({1..$4})bc<<<"$1*$m^2+$m*$2+$3"

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Similar to my previous solution. M, N, C, V are passed as arguments $1 $2 $3 $4.

brainfuck, 87 bytes

+>>,>>,>,>,[-<<[-<+>>+<]<[->+<]<[-<+<[-<+>>>>+<<<]<[->+<]>>>]<[->+<]<++>>>[->>+<<]>>.>]

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Well, since I already made a program for the first part of this challenge, I might as well try to do the second one. Obviously, like the previous answer, this only works with integers and input/output is done in binary (non-competing answer). This code is a bit longer than the previous one because it requires at least one multiplication. However, this is still something Brainfuck can manage, since: $$y \times x^2 = y \times (1 + 3 + 5 + \cdots) = y \times 1 + y \times 3 + y \times 5 + \cdots$$

So the only thing that's required is to use an accumulator starting with 1, calculate the product between the accumulator and M, add to C, and finally increment the accumulator by 2 after each loop. Here's a quick explanation of the code:

+>>,>>,>,>,                              set up (tempAcc=0 Acc=1 tempM=0 M tempSum=0 N C V)
           [                             while V != 0:
            -<<[-<+>>+<]<[->+<]              same as part 1
            <[-<+<                           while M != 0:
                  [-<+>>>>+<<<]<[->+<]           tempSum = Acc * M
            >>>]<[->+<]<++>                  add 2 to Acc
>>[->>+<<]>>.>]                              add tempSum to C; print(C)

I think that's pretty much it, since the third part would be way harder to do with Brainfuck.

Python 3, 52, 50 bytes

lambda M,N,C,V:[M*A*A+N*A+C for A in range(1,V+1)]

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Thanks to Unrelated String for saving 2 bytes.

AWK, 34 bytes

{for(;i++<$4;)print$1*i^2+$2*i+$3}

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R, 28 bytes

Takes M, N, and C as a vector "b".

\(b,v)outer(1:v,2:0,`^`)%*%b

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R, 27 bytes

\(M,N,C,V,x=1:V)M*x^2+N*x+C

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APL+WIN, 17 8 bytes

Thanks to Adam for a saving of 9 bytes.

Prompts for a vector of coefficients matching the order of the polynomial followed by the number of points.

(⍳⎕)⊥¨⊂⎕

Try it online! Thanks to Dyalog Classic

Charcoal, 7 bytes

IEθ↨⊕ιη

Try it online! Link is to verbose version of code. Takes input in JSON format as V and a list of M, N, C. Explanation:

  θ     Input `V`
 E      Map over implicit range
      η Input list `M, N, C`
   ↨    Evaluate polynomial
     ι  Current value
    ⊕   Incremented
I       Cast to string
        Implicitly print

PowerShell, 46 bytes

iex('1..{3}|%{{{0}*$_*$_+{1}*$_+{2}}}'-f$args)

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Basically the same solution as in the earlier post: it's shorter to use the-f format operator to generate a string that is than invoked (iex) as PowerShell command, than to write out the loop directly.

05AB1E, 2 bytes

This challenge made me realize a golf on my answer for the previous challenge, so both are now exactly the same.

Two loose inputs in the order and format \$V\$ and \$[M,N,C]\$.

Try it online or verify all test cases. (Test cases taken from @Adám's APL answer.)

Explanation:

L   # Push a list in the range [1, first (implicit) input V]
 β  # Convert the second (implicit) input-triplet [M,N,C] from a base-v list to
    # base-10 integers for each v in the [1,V]-ranged list
    # (after which this list is output implicitly)

JavaScript (Node.js), 42 bytes

(m,n,c)=>g=v=>v?[...g(v-1),m*v*v+n*v+c]:[]

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Linear solution is 1 bytes longer:

(m,n,c)=>g=v=>v?[c+=n+=m,...g(v-1,n+=m)]:[]

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APL (Dyalog 20.0), 5 bytes

Anonymous tacit infix function, taking V as left argument and [M,N,C] as right argument.

⍪⍤⍳⍛⊥

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⍛⊥ evaluate in mixed-base, but first pre-process the left argument as follows:

 …⍤⍳ expanding to the indices 1…V, and then:

   transform into a column vector (matches up the entire [M,N,C] with each of 1…V)

Vyxal 3, 10 bytes

ɾ:²?×$?×++

taking heavy inspiration from pacman's answer on the previous post

ɾ:²?×$?×++­⁡​‎‎⁡⁠⁡‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁢‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁣‏⁠‎⁡⁠⁤‏⁠‎⁡⁠⁢⁡‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁢⁢‏⁠‎⁡⁠⁢⁣‏⁠‎⁡⁠⁢⁤‏‏​⁡⁠⁡‌⁢⁡​‎‎⁡⁠⁣⁡‏‏​⁡⁠⁡‌⁢⁢​‎‎⁡⁠⁣⁢‏‏​⁡⁠⁡‌⁢⁣​‎‏​⁢⁠⁡‌­
ɾ           # ‎⁡range 1..V
 :          # ‎⁢duplicate the range
  ²?×       # ‎⁣square and multiply by M
     $?×    # ‎⁤fetch other copy of the range and multiply by N
        +   # ‎⁢⁡add MX²+NX
         +  # ‎⁢⁢add C
# ‎⁢⁣implicit output
💎

Created with the help of Luminespire.

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