finding rational zero`s?

Question

my teacher taught us how to do it in a graphing calculator and then after that you do synthethic divisions.
help please i want to know how to do it and do the hw without copying the back of the book for the answers
please and thanks

Answer 1

Suppose you have 2x^3 - 6x^2 + x + 6 = 0
The possible rational roots have a factor of 6 on top and a factor of 2 on bottom. So 6/1, 3/1, 2/1, 1/1, 6/2, 3/2, 2/2, 1/2 plus or minus. You have to guess a root and try it (by synthetic division) until you get one that has a remainder of zero.

Here 2 is a root, so (x-2) is a factor. Do the synthetic division to get the other factor. It's 2x^2 - 2x - 3. If this is factorable you do it; if not leave it.

If you have more specific questions send an IM or email

Answer 2

for all functions, consider f(x) as the numerator
g(x) as the denominator

solve the numerator = 0
solve the denominator = 0

the roots are at x such that f(x) = 0
the vertical asymptotes are at x such that g(x) = 0

if the equations are algebraic
compare the leading exponent of ax^m + ..../ bx^n + ....
if m>n, there is no vertical asymptote
if m= n, there is a vertical asymptote at y = a/b
if m< n, there is a vertical asymptote at y = 0

however, if f(x) and g(x) equal zero at the same value(s) of x....
there is a "problem"....this is called an "indeterminate" form


if you know how to take the derivative....
" l'hopital's rule "
find f' and g'...substitute the indeterminate value, the answer is the value for the "hole" in the equation....
unless you get a 0 in the denominator...in which case, there is a vertical asymptote at that point...


one more "refinement" for making a graph
the space between two vertical asymptotes only has two possible shapes...
if there is a root ....then the graph looks like a tangent or cotangent going through the root
if there is no root...then the graph looks like a "u" or a "n"...it cannot cross the x axis....so, it is bound between the asymptotes..simple substitution will give you the correct shape..
pick a value between the asymptotes...a "+" value means a "u" shape and conversely a "- " value gives a "n"




I hope this helps