Let f(x) = log2 (x+2) +1. Determine the domain, vertical asymptote, and x-intercept of f(x).?

Question

Please show your work :)

Answer 1

let y = f(x) = log2(x+2) + 1.

This function returns proper values only of the argument of the logarithm is positive. Thus, the domain of the function is (-2,infinity).

The vertical asymptote is defined for a rational function, when the denominator tends to zero. For this, rewrite y as

y = log2((x+2)*2) = -log2(1/(2*(x+2))). Thus, we see that the vertical asymptote is the line x = -2.

The x-intercept of a function is the value of x for which the function is zero.

thus 0 = log2(2*(x+2)) which implies that x+2 = 1/2 or x = -3/2.

Answer 2

Since log2(0) is undefined, the domain is - 2 < x
Vertical asymptote is then x = - 2.
2^(y - 1) = x + 2
x = 2^(y - 1) - 2
x = 2^(0 - 1) - 2
x = - 1.5