precalculus

Precalculus Assignment
1. Solving, log2x+ log4x = 6⇒ log2x+ 12 log2x = 6⇒ log2x = 4⇒ x = 2
4 = 16.
2. For function with vertical asymptote at x=3, then the denominator must have a zero at 3 or it must contain (x-3) term. For
function to have x intercepts(y=0) at x=0, x=7, the function must be of form f(x) = x(x−7)×G(x) atleast because in such
a case, we see that when we substitute the values 0 ot 7, we get f(x)=y=0 which is precisely what an X intercept is. Further,
for function to have a horizontal asymptote at y=5, coefficient of leading term in numerator must have coefficient 5. So, we
have a sample rational function which satisfies the above requirements(in it’s simplest form G(x)=1) as, f(x) = 5x(x−7)(x−3)(x−8) .
Clearly, as x → ∞, Limx→∞f(x) = 5 ⇒ Indeed the horizontal asymptote is y = f(x) = 5. Also, it has X intercepts as
asked for and also, Limx→3f(x) = ∞. However, also note that the function now...