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 points a and d belong to the x−axis and m∠bad=60°

The value of a = ±(√3)/(6)

Step-by-step explanation:

Points A and D belong to the x−axis.

All vertices on the parabola y = a (x+1)(x−5) = a (x² - 4x - 5)

So, points A and D represents the x-intercept of the parabola y

To find x-intercept, put y = 0

∴ a (x+1)(x−5) = 0  ⇒ divide both sides by a

∴ (x+1)(x−5) = 0 ⇒ x = -1 or x = 5

so, the x-coordinate of Point A is -1 or 5

And given that: m∠BAD=60°

So, the tangential line of the parabola at point A has a slope of 60°

∴ y' = tan 60° = √3

∴ y' = a (2x-4)

∴ a (2x-4) = √3

∴ a = (√3)/(2x-4)

Substitute with x = -1 ⇒ a = (√3)/(-6)

Substitute with x = 5 ⇒ a = (√3)/(6)

So, The value of a = ±(√3)/(6)

Also, see the attached figure, it represents the problem in case of a = (√3)/(-6)

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

4444444444444 i think