Respuesta :
Answer:
Option: C is correct.
[tex]y=7\times (0.5)^x[/tex]
Step-by-step explanation:
We are given that when x increases from 'a' to 'a+2' then y must increase by a factor of 1/4=0.25.
i.e. when x=a and x'=a+2.
then [tex]\dfrac{y'}{y}=0.25[/tex] where y' is the function after putting x' to the old function.
A)
[tex]y=3\times 2^x[/tex]
when x=a
[tex]y=3\times 2^a[/tex]
when x'=a+2
[tex]y'=3\times 2^{a+2}\\\\y'=3\times 2^a\times 2^2\\\\y'=3\times 2^a\times 4\\\\y'=4\times y\\\\\dfrac{y'}{y}=4\neq \dfrac{1}{4}[/tex]
Hence, option (A) is incorrect.
B)
[tex]y=2x+5[/tex]
when x=a
[tex]y=2a+5[/tex]
when x'=a+2
[tex]y'=2(a+2)+5\\\\y'=2a+4+5\\\\\\y'=y+4[/tex]
Here we do not get a factor of [tex]\dfrac{1}{4}[/tex].
Hence, option B is incorrect.
C)
[tex]y=7\times (0.5)^x[/tex]
when x=a
[tex]y=7\times (0.5)^a[/tex]
when x'=a+2
[tex]y=7\times (0.5)^{a+2}\\\\y=7\times (0.5)^a\times (0.5)^2\\\\y=y\times 0.25\\\\\dfrac{y'}{y}=0.25[/tex]
Hence we get a factor of [tex]\dfrac{1}{4}=0.25[/tex]
Hence, option C is correct.
D)
[tex]y=\dfrac{1}{2}x-4[/tex]
when x=a
[tex]y=\dfrac{1}{2}a-4[/tex]
when x'=a+2
[tex]y'=\dfrac{1}{2}(a+2)-4\\\\y'=\dfrac{1}{2}a+1-4\\\\y'=y+1[/tex]
Here also we did not get a factor of [tex]\dfrac{1}{4}[/tex].
Hence, option D is incorrect.
Hence, the function is:
[tex]y=7\times (0.5)^x[/tex]
Hence, option C is correct.
