Rational Functions -- Section 5.6 -- Find the vertical and horizontal asymptotes, if any.  Solve Step VA: Set the denominator equal to zero. Set x - 3 = 0   So, x  = 3. Obviously, f(3) = undefined.  What happens as x approaches 3?  i.e. x → 3 From the left hand side of 3 From the right hand side of 3 The Vertical Asymptote, VA, is the line, x  = 3.  From the sketch of the graph we see y = 0 is the horizontal asymptote.  Look what happens when we plug in large and small numbers.  The Horizontal Asymptote, HA, is the line, y  = 0 Whenever the degree of the numerator is less than the degree of the denominator, the HA is y = 0. As x →∞ (or -∞) , f (x) → 0 because the denominator will increase much faster than the numerator.  Find the Vertical Asymptotes, if any.  [Solution]  Find the vertical and horizontal asymptotes, if any.  Then find x and y intercepts, if any. Solve Step Vertical Asymptote:  x  - 2  = 0, x  = 2 The vertical asymptote is the line x = 2 Horizontal Asymptote:  y = 3 The horizontal asymptote is the line y = 3  If the degree of the numerator and the denominator are the same, as x approaches ±∞ , then y approaches the coefficients of the of the leading term in  the numerator and denominator. x - intercept: (-4/3, 0) Set y = g(x) = 0. The only way a fraction can equal zero is if the numerator is zero.  y - intercept: (0, -2) g(0) = -2 Set x  = 0. Let's put it all together in a graph. Notice the x and y intercepts are denoted by the blue dots.  The asymptotes are denoted by the broken blue lines.  Find the vertical and horizontal asymptotes, if any.  Then find x and y intercepts, if any.  [Solution]  Is there really an Asymptote? Solve Step Property Vertical Asymptote: None VA: Set the denominator equal to zero. Is this really an asymptote?? Something is wrong. As x →4,  f→6.   This is not an asymptote but a hole in the function at x = 4.  After we reduce the fraction, we get f(4) = 6,but x ≠ 4 in original function.  This means there is an hole at (4,6). Horizontal Asymptote:  None as  x → ∞, f (x) = x + 2 → ∞ as  x → -∞, f (x) = x + 2 → -∞ Since the factor (x - 4) cancels out, we are left with the line y  = x + 2 where x ≠ 4.   Is there really an Asymptote?  [Solution] Tutorials and Applets by
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