Here's the first question. read http://www.mathhelpforum.com/math-he...-help-plz.html for info on vertical and horizontal asymptotes.

A)Find the domain of f(x)

the domain of a function is the set of all inputs (in this case x-values) for which the function is defined. it is almost always easier to find the x's for which the function is NOT defined, and say it's defined for all x's but those. so let's do that.

this is a rational function, it is defined everywhere as long as it's denominator is not zero.

so we need to find where x^2-2x-15 is zero.

x^2-2x-15 = 0

=> (x - 5)(x + 3) = 0

=> x = 5 and x = -3

these are the x's that make the denominator zero, so the domain is given by:

dom(f) = {x : x not= 5 and x not= -3}

or in interval form:

dom(f) = (-infinity, -3)U(-3,5)U(5, infinity)

B) Find the x and y intercepts for f(x)

for the x-intercepts, set y = 0

=> 0 = (x-5)/(x^2-2x-15)

=> 0 = x - 5 ...........since x^2-2x-15 can't be zero

=> x = 5 ..........x-intercept

for the y-intercept, set x = 0

=> y = (0-5)/[0^2-2(0)-15] = -5/-15 = 1/3 ........y-intercept

C) find the vertical and horizontal asymtotes for f(x)

vertical asymptotes occur at x-values for which the function is undefined, so it will be all x's NOT in the domain. we did the work in the first part so i won't do it again.

our vertical asymptotes are: x = 5 and x = -3

for the horizontal asymptote we have y = 0, since the degree of the numerator is smaller than the degree of the denominator. the official way to do this question is to take the limit as x goes to infinity and -infinity, if the result is a number, that's the horizontal asymptote

D) find the derivitive for f(X) and the critical points

ok, this is a rational function, let's use the quotient rule to find the derivative.

Recall, Quotient rule:

(u/v)' = (v*u' - u*v')/v^2

f(x) = (x-5)/(x^2-2x-15)

By the Quotient rule:

=> f ' (x) = [(x^2 - 2x - 15) - (x - 5)(2x - 2)]/[(x^2-2x-15)^2]

.............= (x^2 - 3x -10)/[(x^2-2x-15)^2]

...we can factorize and simplify this a bit, but i won't bother

critical points occur when the derivative is zero.

for critical points, set f ' (x) = 0

=> (x^2 - 3x -10)/[(x^2-2x-15)^2] = 0

=> x^2 - 3x - 10 = 0

=> (x - 5)(x + 2) = 0

=> x = 5, x = -2

x = 5 is an asymptote, sox = -2 is the only critical point.(actually, the (x - 5) would cancel if we simplified f ' (x), so don't worry about it too much)

E) find the second derivative and the inflection points of F(X)for the second derivative, we again use the quotient rule, but this time on the first derivative.

f '' (x) = (f ' (x))' = [(x^2 - 3x -10)/[(x^2-2x-15)^2]]'

.......................= {[(x^2-2x-15)^2]*(2x - 3) - (x^2 - 3x - 10)(2*(x^2 - 2x - 15)*(2x - 2)}/[[(x^2-2x-15)^4]

now i really don't want to simplify that!

for the POSSIBLE inflection points, set f '' (x) = 0

=> {[(x^2-2x-15)^2]*(2x - 3) - (x^2 - 3x - 10)(2*(x^2 - 2x - 15)*(2x - 2)}/[[(x^2-2x-15)^4] = 0

=> [(x^2-2x-15)^2]*(2x - 3) - (x^2 - 3x - 10)(2*(x^2 - 2x - 15)*(2x - 2) = 0

=> (x - 5)^2*(x + 3)^2*(2x - 3) - 2(x - 5)(x + 2)(x - 5)(x + 3)(2x - 2) = 0

=> [(x - 5)^2(x + 3)]*[(x + 3)(2x - 3) - 2(x + 2)(2x - 2)] = 0

=> (x - 5)^2(x + 3) = 0

=>x = 5, x = -3

or

=> (x + 3)(2x - 3) - 2(x + 2)(2x - 2) = 0

=> 2x^2 + 3x - 9 - (4x^2 + 4x - 8) = 0

=> 2x^2 + 3x - 9 - 4x^2 - 4x + 8 = 0

=> -2x^2 - x - 1 = 0

=> 2x^2 + x + 1 = 0

By the quadratic formula:

x = [-1 +/- sqrt(1 - 8)]/4

no real solution----> there are no inflection points

EDIT: it turns out that x = 5 is not an asymptote either. Since we could have factorized the bottom and cancel the (x - 5) on top. so the domain includes 5 and 5 is not a critical point. In retrospect, life would have been a LOT easier if we factorized and simplified before doing all these operations. now that youknow the steps, factorize f(x) cancel what needs to be canceled and redo the steps yourself.