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In which direction does the parabola x+3y2=−4x + 3y^2 = -4 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=−4x + 3y^2 = -4 open?\newlineChoices:\newline(A) up\newline(B) down\newline(C) right\newline(D) left
  1. Identify Parabola Type: Look at the equation x+3y2=−4x + 3y^2 = -4 and identify the squared term to determine if the parabola is vertical or horizontal.\newlineSince yy is squared, it's a horizontal parabola.
  2. Rearrange Equation: Rearrange the equation to solve for xx.x=−3y2−4x = -3y^2 - 4
  3. Determine Parabola Direction: Check the coefficient of y2y^2 to determine the direction of the parabola.\newlineThe coefficient is −3-3, which is less than 00, so the parabola opens to the left.

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