Select the correct answer.
Consider the function RX) = x and the function g(x) shown below. How will the graph of g(x) differ from the graph of Rx)?
g(x) = px - 2) = (x - 2)
ОА
The graph of g(x) is the graph of Rx) shifted up 2 units.
ОВ. The graph of g(x) is the graph of Rx) shifted to the right 2 units.
OC. The graph of g(x) is the graph of Rx) shifted down 2 units.
OD. The graph of g(x) is the graph of Ax) shifted to the left 2 units.

Select the correct answer Consider the function RX x and the function gx shown below How will the graph of gx differ from the graph of Rx gx px 2 x 2 ОА The gra class=

Respuesta :

Answer:

B

Step-by-step explanation:

just think about it :

can it move up or down ? no, because for a specific input value still the same functional result is calculated (nothing is getting bigger or smaller).

all that is happening that way is that now, with using g(x), the original f(x) functional values happen now 2 units "later" = to the right (if you consider the x-axis a time line growing to the right). we are getting the functional value of f(x-2) at x and not at x-2 for g(x).

for example

the functional values are for x² (just some integers to make it easier) :

x = 1, 2, 3, 4, 5, ...

getting

f(1), f(2), f(3), f(4), f(5), ...

leading to

1², 2², 3² 4², 5², ...

which is

1, 4, 9, 16, 25, ...

now, let's say we start looking at x = 3

x = 3, 4, 5, 6, 7, ...

getting

g(3), g(4), g(5), g(6), g(7), ..

leading to

1², 2², 3² 4², 5², ...

which is

1, 4, 9, 16, 25, ...

so, now we are getting the functional value at e.g. x = 5 that we got originally for x = 3 (9).

therefore, under g(x) the original functional values still "happen", they just simply "happen" 2 units "later" (to the right).

in the same way

g(x) = f(x+2) moves everything 2 units to the left (now things are happening "earlier").