Given the vertices of isosceles trapezoid abcd: a(2, 0), b(4, 0), c(5, -2), d(1, -2). if the image is reflected over the y-axis, c' = (5, 2) (-5, 2) (-5, -2) (5, -2) here is the image

C' = (-5, -2)

Step-by-step explanation:

Reflection over the y-axis negates the x-coordinate.

C(5, -2) ⇒ C'(-5, -2)

The correct option is 3.

Step-by-step explanation:

The vertices of isosceles trapezoid ABCD are A(2, 0), B(4, 0), C(5, -2), D(1, -2).

If a figure reflected over the y-axis, then the relation between coordinates of preimage and image is defined as

It is given that ABCD reflected over the y-axis and coordinates of point C are (5,-2), so the coordinates of C' are

The coordinates of C' are(-5,-2). Therefore the correct option is 3.

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

u bout ugly as   a mf

step-by-step explanation: lol