How do you graph #(x^3-6x+5)/(x^2-3)#?

1 Answer
Jan 26, 2016

Answer:

graph{(x^3 - 6x+5)/(x^2-3) [-7, 10.866, -3.71, 5.22]}

Explanation:

Let #f(x) = (P(x))/(Q(x)) "where" P(x) = x^3 - 6x +5 " and " Q(x) = x^2 - 3#

Find the y-intercept ==> #F(0) = -5/3#
Find the x-intercept by setting the nominator to zero and solving for x. #0 = x^3 - 6x +5 #
#x_1 = 1; -1/2 +1/2sqrt(21); -1/2 - 1/2sqrt(21)#
y-asymptote ==> set denominator to zero and solve for x
#x^2 -3 =0; x_(1,2)= +-sqrt(3) #
x- asymptot since since the power of P(x) is > Q(x)
Oblique asymptote is by long division ==> y = x