HOME | DD
CubedCake β Undertale Rose Collection - Frisk
#frisk #undertale #undertalegame #undertale_game #undertalefrisk #friskundertale #frisk_undertale #undertale_frisk
Published: 2015-10-31 17:43:43 +0000 UTC; Views: 37185; Favourites: 1476; Downloads: 0
Redirect to original
Description
This collection is about undertale characters relating to the color of rosesFrisk - Red Rose: love, courage, and passion
Β
Rose Collection completed!
Sans |Papyrus |Undyne |Alphys |Mettaton |Toriel |Asgore |Asriel |Chara
If you also like to share, check out the other media links!
Posted on tumblr
Posted on twitter
Related content
Comments: 33
π: 0 β©: 0
May I use this picture for a Undertale music video? I'll give credit.Β
π: 0 β©: 0
This is so adorable! Would you mind if I used this in a cover I'm doing?
π: 0 β©: 1
As long my art is credited somewhere in the description and non-profited, then it's fine with me!
π: 0 β©: 1
The color red in the game means DETERMINATION. But, In rose color theme thing, it does not.
π: 0 β©: 0
Excuse me, could I repost your artwork on a fan made Instagram page based on Frisk X Sans? Your artwork will be credited to you. If you have any questions send me a private message! βΊοΈ
π: 0 β©: 1
Well since you asked first before reposting, I'm ok with it. Please credit me with my instagram account. It's cubedcakeΒ @Β instagram.
π: 0 β©: 1
Kinda getting an Ib vibe from this . Love it
π: 0 β©: 0
Frisk, aka the Savior of the Underground, the "silent but loving" type. Β Red roses are my favorite, not only because they look and smell nice but because of what they represent. Β I didn't know they could represent courage too. Β I thought they meant love, approval and passion only.
π: 0 β©: 1
Red roses can represent in many ways since it's a very common flower. Though most of the information that I research on red roses is mostly define with those words. I mean it does take a lot of guts and courage to confess your love right? But red roses can also relate to approval too.
π: 0 β©: 0
as long as you credit me, it's fine!
π: 0 β©: 0
