What set of reflections would carry trapezoid abcd onto itself? trapezoid abcd is shown. a is at 1, negative 1. b is at 2, negative 2. c is at 3, negative 2. d is at 4, negative 1. y=x, x-axis, y=x, y-axis x-axis, y=x, x-axis, y=x y-axis, x-axis, y-axis x-axis, y-axis, y-axis

The correct answer is:

y=x, x-axis, y=x, y-axis

Explanation:

A reflection across the line y=x maps every point (x, y) to (y, x); it switches the coordinates but does not negate them.

This means for A(1, -1), we would have A'(-1, 1); B(2, -2)→B'(-2, 2); C(3, -2)→C'(-2, 3); and D(4, -1)→D'(-1, 4).

A reflection across the x-axis negates the y-coordinate; algebraically,

(x, y)→(-x, y).

This takes our new points A'(-1, 1)→A''(-1, -1); B'(-2, 2)→B''(-2, -2); C'(-2, 3)→C''(-2, -3); and D'(-1, 4)→D''(-1, -4).

Reflecting again across the line y=x will again switch the x- and y-coordinates:

A''(-1, -1)→A'''(-1, -1); B''(-2, -2)→B'''(-2, -2); C''(-2, -3)→C'''(-3, -2); and D''(-1, -4)→D'''(-4, -1).

Reflecting across the y-axis will negate the x-coordinate:

A'''(-1, -1)→(1, -1); B'''(-2, -2)→(2, -2); C'''(-3, -2)→(3, -2); and D'''(-4, -1)→(4, -1).

These are the same as our original points.

it's both, now can you answer my most recent question?

answer: yes, since the line has a steeper rate of change than the exponential the y-values will be greater and they will only intersect one time.