Question

Let f (x) = 4X^2 -9 /X^2-16

(a) State the y-intercept

. (b) State the x-intercept(s).

(c) State the vertical asymptote(s).

(d) State the horizontal asymptote.

Answer 1

A) For y-intercept, take x = 0

y = (0 - 9)/(0 - 16) = 9/16

So the y-intercept is (0, 9/16)

B) For x-intercept, take y = 0

(4x^2 - 9)/(x^2 - 16) = 0

Or, 4x^2 - 9 = 0

Or, x^2 = 9/4

Or, x = 3/2

So the x-intercept is (3/2, 0)

C) For vertical asymptote, take denominator = 0

X^2 - 16 = 0

Or, X^2 = 16

Or, X = +/- 4

So the vertical asymptote is at x = -4 and x = 4

D) Since degrees are the same, divide the leading coefficient of the numerator and denominator = 4/1 = 4

So the horizontal symptote is at y = 4