Question

Graph the equation ?? = ??^2????6/2??^2?2 by hand. find and show the algebra to find the EQUATIONS of the vertical asymptote(s), horizontal asymptote (if there is one), as well as the x and y intercepts.

Answer 1

Given equation is : y = (x²-x-6)/(2x²-2)

Graph :

For vertical asymptote we have to equate denominator with 0.

Then, 2x²-2 = 0

i.e., x²-1 = 0

i.e., x² = 1

i.e., x = -1, 1

Therefore, vertical asymptotes are : x = -1 and x = 1.

Here the highest power of x in both numerator and denominator are same.

Then, the horizontal asymptote is : y = (coefficient of x² in numerator)/(coefficient of x² in denominator)

i.e., y = 1/2

Therefore, the horizontal asymptote is : y = 1/2.

For x intercepts, we have to solve y = 0.

i.e.,  (x²-x-6)/(2x²-2) = 0

i.e.,  x²-x-6 = 0

i.e., (x-3)(x+2) = 0

i.e., x = 3, -2

Therefore, x intercepts are (3,0) and (-2,0).

For y intercepts we have to put x = 0 in y = (x²-x-6)/(2x²-2).

Then we get, y = (-6)/(-2) ,i.e., y = 3

Therefore, y intercept is (0,3).