Question

Log based question, i just need help with parts b and c

Answer 1

b) There is translation 3 units to the right and 1 unit up

c) The domain is {x I x > 3}

The equation of the asymptote is x = 3

Explanation:

* Lets revise the rule of the translation

- If the function f(x) translated horizontally to the right

by h units, then the new function g(x) = f(x - h)

- If the function f(x) translated horizontally to the left

by h units, then the new function g(x) = f(x + h)

- If the function f(x) translated vertically up

by k units, then the new function g(x) = f(x) + k

- If the function f(x) translated vertically down

by k units, then the new function g(x) = f(x) – k

* Now lets solve the problem

b) ∵ f(x) =
`log_(2) (x-3)+1`

∵ The parent function is
`log_(2) x`

∴ x is changed to (x - 3), that means there is a translation 3 units

to the right

∵ We add the parent function by 1, that means there is a translation

1 unit up

* There is translation 3 units to the right and 1 unit up

c) To find the domain of the function, find the values of x which

make the function undefined

∵
`log_(2)(0)` is undefined

∴ x - 3 can not be 0

∵ x - 3 = 0 ⇒ add 3 to both sides

∴ x = 3

∴ The domain of the function is all real number greater than 3

* The domain is {x I x > 3}

∵ x can not be 3

∴ There is a vertical asymptote, its equation is x = 3

* The equation of the asymptote is x = 3

# Look to the attached graph for more understand for the domain

and the equation of the asymptote