Obviously I cannot help you with the graphing, but
I assume you can do the reflection about the x-axis
(any point on the x-axis would stay, all points keep their x's, and their y's become opposite.
e.g. A(6,8) ---> A1(6,-8)
repeat for the other points and graph the new figure, then shade as instructed.
y = -14 is simply a horizontal line 14 units below the x-axis (two points on it would be (5,-14) and (-5,-14) , join them)
Now you are reflecting all points around that line
the x's would stay the same, but for the y, calculate how far above -14 it is, then move down just as far below the -14
e.g. B(12,4) is 18 units above the line y = -14
(how did I get 18 ??)
so it moves 18 units below that line, as a result:
B(12,4) ---> B2(12,-32)
you can check your new points are correct by taking the average of your y values of the old point and the new point, you should get -14
(-32 + 4)/2 = -14
give it a try, I suggest you print out my instructions.
Given a quadrilateral having coordinates A(6,8), B(12,4), C(6,0) and D(4,0), first graph the quadrilateral on graph paper and connect the vertices. Shade in the quadrilateral and then reflect it across the x-axis. Do NOT shade in the resulting image. Now reflect this new image across the line y=-14 and shade in the final image. Explain what the same is and what is different about the two shaded quadrilaterals(the original one and the final one). What is the relationship between the two lines of reflection? Do you think that makes a difference? Why or why not? Please help me with this and thank you to helps me.
Can someone help me with this? I don't understand how to do the y=-14 part my teacher didn't explain it to us today.
2 answers
I don't get it