The coordinates of the endpoints of `bar(ab)` and `bar(cd)` are a(x1, y1), b(x2, y2), c(x3, y3), and d(x4, y4). which condition proves that ` bar(ab)||bar(cd)`? a. `((y_4-y_2)/(x_4-x_2)= (y_3-y_1)/(x_3-x_1))` b. `((y_4-y_3)/(y_2-x_1)= (x_4-x_3)/(x_2-x_1))` c. `((y_4-y_3)/(x_4-x_3)= (y_2-y_1)/(x_2-x_1))` d. `((y_2-y_1)/(x_4-x_3)= (x_2-x_1)/(y_4-y_3))`

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if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

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