Cracking The Ap Calculus ab Exam 2018

vertical    asymptote   at  x   =   3.  There   is  no  horizontal  asymptote,  but notice  that    the degree  of  the

numerator   of  the function    is  1   greater than    the denominator.    This    means   that    there   is  an  oblique

(slant) asymptote.  We  find    this    by  dividing    the denominator into    the numerator   and looking at

the quotient.   We  get

This means that as x → ±∞, the function will behave like the function y = x + 3. This means that

there    is  an  oblique     asymptote   of  y   =   x   +   3.  Next,   we  take    the     derivative:

=    =   .   Next,   we  set     the

derivative  equal   to  zero    to  find    the critical    points. There   are two solutions:  x   =   3   +       and x   =

3   −    .  We  plug    these   values  into    the original    equation    to  find    the y-coordinates   of  the critical

points:  When   x    =   3   +   ,   =   6   +   2 .     When   x    =   3   −   ,  y    =  

    =   6   −2 .    Thus,   we  have    critical    points  at  (3  +    ,  6   +   2 ) and (3  −    ,

6   −   2 ).    Next,   we  take    the second  derivative:  .  If  we  set this    equal   to  zero,

there    is  no  solution.   Therefore,  there   is  no  point   of  inflection.     But,    notice  that    the     second

derivative  changes sign    from    negative    to  positive    at  x   =   3.  This    means   that    the curve   is  concave

down    for values  of  x   less    than    x   =   3   and concave up  for values  of  x   greater than    x   =   3.  Next,   we

need    to  determine   if  each    critical    point   is  maximum,    minimum,    or  something   else.   If  we  plug    x

=    3   +   into    the     second  derivative,     the     value   is  positive,   so  (3  +   ,   6   +   2 )     is  a

minimum.    If  we  plug    x   =   3   −       into    the second  derivative, the value   is  positive,   so  (3  −    ,

6   −   2 ) is  a   maximum.