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In which direction does the parabola x+3y2=3x + 3y^2 = 3 open?\newlineChoices:\newline(A)up\newline(B)down\newline(C)right\newline(D)left

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Q. In which direction does the parabola x+3y2=3x + 3y^2 = 3 open?\newlineChoices:\newline(A)up\newline(B)down\newline(C)right\newline(D)left
  1. Isolate xx: Now, we gotta get the equation in the standard form for a parabola.\newlineSo, we'll isolate xx by subtracting 3y23y^2 from both sides.\newlinex+3y2−3y2=3−3y2x + 3y^2 - 3y^2 = 3 - 3y^2\newlinex=−3y2+3x = -3y^2 + 3
  2. Determine opening direction: Alright, let's figure out which way this thing opens.\newlineThe coefficient of y2y^2 is −3-3, which means a=−3a = -3.\newlineSince aa is less than 00, the parabola opens to the left.

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