+12 a parabola curv have an equation as 3x + 1 = y^2 -y -1 which going through the y-axis in the positive side and also contain the smallest x value on the curv. Use quadratic equation to solve the point where the smallest x value is (coordinate).
+129842 a parabola curv have an equation as 3x + 1 = y^2 -y -1 which going through the y-axis in the positive side and also contain the smallest x value on the curv. Use quadratic equation to solve the point where the smallest x value is (coordinate).
The quadratic formula probably isn't necessary here....let's simplify the equation
3x + 1 = y^2 -y -1 subtract 1 from both sides
3x = y^2 - y - 2 divide both sides by 3
x = (1/3)y^2 - (1/3)y - 2/3 complete the square on y ....add 2/3 to both sides
x + 2/3 + 1/12 = (1/3) (y^2 - y + 1/4) factor the right side, simplify the left
(x + 3/4) = (1/3) (y -1/2)^2 x takes on its minimum value at the vertex which is (-3/4, 1/2).....so x is minimum at -3/4
Here's the graph : https://www.desmos.com/calculator/j3plekvi77
