Suppose that f(x) = x3 and g(x) = -2x3-7. which statement best compares
the graph of g(x) with the graph of f(x)?
a. the graph of g(x) is the graph of fx) compressed vertically,
flipped over the x-axis, and shifted 7 units to the right.
b. the graph of g(x) is the graph of f(x) stretched vertically, flipped
over the x-axis, and shifted 7 units to the right.
c. the graph of g(x) is the graph of fx) stretched vertically, flipped
over the x-axis, and shifted 7 units down.
d. the graph of g(x) is the graph of fx) compressed vertically,
flipped over the x-axis, and shifted 7 units down.

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

nlgqer

step-by-step