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hjlf
3 years ago
12

Can someone complete the square

Mathematics
1 answer:
pshichka [43]3 years ago
7 0
To do these, start by looking at the "b" value -6.
divide it by 2

-6/2 = -3

now square this number

(-3)^2 = 9

this is what you need for the "c" value

there is only a 5 for the c value so add 4 to both sides of the equation. ( +4 = +4)

y +4 = x^2 -6x +5 +4
y +4 = x^2 -6x +9
y +4 = (x -3)^2
y = (x -3)^2 - 4

vertex ( 3, -4) upwards facing like a bowl, because the "a" value is positive. So the vertex is the minimum, lowest point on the graph.


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6 0
2 years ago
Please help it will help a lot i really need this!!
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Answer:

A' = (0,0)

B'=(8,0)

C'=(8,2)

D'=(0,2)

It is not a rigid motion.

Step-by-step explanation:

To use the mapping rule, substitute the original x and y values in it.

The coordinates of A are (0,0).

Using the mapping rule, the x coordinate of A' = 2x0 = 0

Using the mapping rule, the y coordinate of A' = (1/2)x0=0

So A' will not change locations. The image will be at (0,0).

The coordinates of B are (4,0)

Using the mapping rule, the x-coordinate of B' = 2x4 = 8

Using the mapping rule, the y-coordinate of B' = (1/2)x0=0

Therefore the image of B' will be located at coorindate (8,0)

The coorindates of C are (4,4).

Using the mapping rule, the x-coordinate of C' = 2x4=8

Using the mapping rule, the y-coordinate of C' = (1/2)x4=2

Therefore the image of C' will be located at coordinate (8,2)

The coordinates of D are (0,4)

Using the mapping rule, the x-coordinate of D' = 2x0 = 0

Using the mapping rule, the y-coordinate of D' = (1/2)x4=2

Therefore the image of D' will be located at coordinate (0,2)

<em>Is the transformation a rigid motion?</em>

NO, this transformation is not a rigid motion because the relative distance between the points does not stay the same after they have been transformed. The transformation is not a translation , rotation, reflection, nor glide reflection.

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