SOLUTION: My teacher would like us to graph, including any oblique asymptotes, y= f(x) = X^2+5 ----- x+1 find intercepts, verticle asymptotes, and horizontal asymptotes. I have the grap

Algebra ->  Rational-functions -> SOLUTION: My teacher would like us to graph, including any oblique asymptotes, y= f(x) = X^2+5 ----- x+1 find intercepts, verticle asymptotes, and horizontal asymptotes. I have the grap      Log On


   

Question 112972This question is from textbook college algebra
: My teacher would like us to graph, including any oblique asymptotes, y= f(x) = X^2+5
-----
x+1
find intercepts, verticle asymptotes, and horizontal asymptotes. I have the graph done, I just need to know how to do the written work.
Thank you. This question is from textbook college algebra

Answer by chitra(359) About Me  (Show Source):
You can put this solution on YOUR website!
The given expression is: y = f(x) = +%28x%5E2+%2B+5%29%2F%28x+%2B+1%29

The intercepts are given as follows:

Y - intercept put x = 0

==> y = (0 + 5)/(0 + 1)

==> y = 5/1

==> y = 5

The x-intercepts are given when y = 0

==> 0 = %28x%5E2+%2B+5%29%2F%28x+%2B+1%29

==> 0 = x%5E2+%2B+5

==> x^2 = - 5

==> +x+=+i%28sqrt%285%29%29 (+ -)


The graph looks this way:

+graph%28+300%2C+300%2C+-15%2C+15%2C+-15%2C+15%2C+%28x%5E2+%2B+5%29%2F%28x+%2B+1%29%29+

The vertical and the horizontal asymptotes are found as follows:


The vertical asymptotes come from the zeroes of the denominator, so I'll set the denominator equal to zero and solve.

x + 1 = 0

==> x = -1

Thus the vertical asymptotes.

Since the degree of the numerator is one greater than the degree of the denominator, I'll have a slant asymptote (not a horizontal one), and I'll find the slant asymptote by long division.

Thus by long dision we find that y = x + 1 is the slant asymptote for the given equation.