How do you graph rational functions?

1 Answer
Jul 12, 2015

Answer:

See explanation...

Explanation:

Suppose #f(x) = g(x)/(h(x)) = (a_nx^n+a_(n-1)x^(n-1)+..+a_0)/(b_mx^m+b_(m-1)x^(m-1)+...+b_0)#

If #g(x)# and #h(x)# have some common factor #k(x)# then let

#g_1(x) = g(x)/(k(x))# and #h_1(x) = g(x)/(k(x))#.

The graph of #g_1(x)/(h_1(x))# will be the same as the graph of #g(x)/(h(x))#,

except that any #x# where #k(x) = 0# is an excluded value.

Assuming #g(x)# and #h(x)# have no common factor, then there will be vertical asymptotes wherever #h(x) = 0#. If a root is not a repeated root (or is repeated an odd number of times) then the limit on one side of the asymptote will be #oo# and on the other #-oo#. If the root has even multiplicity then the limit on both sides of the asymptote will be the same: #oo# or #-oo#.

If #n < m# then #f(x)->0# as #x->+-oo#

If #n >= m# then divide #g(x) /(h(x))# to get a polynomial quotient and remainder. The polynomial quotient is the oblique asymptote as #x->+-oo#.

For example, if #f(x) = (x^3 + 3)/(x^2 + 2)#, then:

#f(x) = (x^3+3)/(x^2+2) = (x^3+2x-2x+3)/(x^2+2) = x - (2x-3)/(x^2+2)#

So the oblique asymptote of #f(x)# is #y = x#

Intercepts with the #x# axis are where #f(x) = 0#, which mean where #g(x) = 0#.

The intercept with the #y# axis is where #x=0#, so just substitute #x=0# into the equation for #f(x)# to find #f(0) = a_0/b_0#

Apart from all this, just pick some #x# values and calculate #f(x)# to give you coordinates #(x, f(x))# through which the graph must pass.