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An isosceles trapezoid ABCD with height 2 units has all its vertices on the parabola y=a(x+1)(x−5). What is the value of a, if p

horrorfan [7]

Answer:

a = -0.3575

Step-by-step explanation:

The points A and D lie on the x-axis, this means that they are the x-intercepts of the parabola, and therefore we can find their location.

The points A and B are located where

$y=a(x+1)(x-5)=0$

This gives

$x=-1$

$y=5$

Now given the coordinates of A, we are in position to find the coordinates of the point B. Point B must have y coordinate of y=2 (because the base of the trapezoid is at y=0), and the x coordinate of B, looking at the figure, must be x coordinate of A plus horizontal distance between A and B, i.e

$B_x=A_x+\frac{2}{tan(60)} =-1+\frac{2\sqrt{3} }{3}$