Write a coordinate proof. Prove quadrilateral WXYZ, having vertices W(-5, 6), X(-1, 10), Y(1, 4), and Z(-2, -3) is a trapezoid.

The diagonals are bisectors of each other and meet at  (1, -5/2) but are not congruent, so WXYZ is a parallelogram and not a rectangle

Step-by-step explanation:

for parallelograms.

the diagonals intersect at the midpoints

W                             X

Z                             Y

Diagonals are  WY  and  XZ

WY :   (-4, -3) to  (6, -2)

midpoint is   [(-4 + 6)/2  ,  (-3 - 2)/2 ] = ( 1, -5/2 )

XZ :   (0, -1)  to  (2, -4)

midpoint is  [ (0 + 2)/2 ,  (-1 + -4)/2 ]  = (1, -5/2)

The diagonals are bisectors of eachother.

Next show that these diagonals are not congruent.

length of  WY  =  root (  (-4 - 6)^2  +  (-3 - (-2))^2 )

WY = root ( 100  +  1)

WY = root(101)

length of XZ  =  root (  (0 - 2)^2  +  ( -1 - -4)^2 )

XZ =  root (4  +  9) = root (13)

we see that  WY  is not equal to XZ

...

so  ..  quadrilateral WXYZ is a parallelogram but not a rectangle.

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

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50