Cracking The Ap Calculus ab Exam 2018

(Marvins-Underground-K-12) #1
is  perpendicular   to  the tangent line,   so  its slope   will    be  the negative    reciprocal  of  the tangent

line’s  slope.  In  this    case,   the slope   of  the normal  line    is   .  Finally,    we  plug    in  the point

(3, 7)  and the slope   m   =       to  get the equation    of  the normal  line:   y   −   7   =    (x −   3).


  1. x = 9


A   line    that    is  parallel    to  the y-axis  has an  infinite    (or undefined)  slope.  In  order   to  find    where

the normal  line    has an  infinite    slope,  we  first   take    the derivative  to  find    the slope   of  the tangent

line:       =   2(x −   9)(1)   =   2x  −   18. Next,   because the normal  line    is  perpendicular   to  the tangent

line,   the slope   of  the normal  line    is  the negative    reciprocal  of  the slope   of  the tangent line:   m   =   

. Now, we need to find where the slope is infinite. This is simply where the


denominator of  the slope   is  zero:   x   =   9.

11.

A   line    that    is  parallel    to  the x-axis  has a   zero    slope.  In  order   to  find    where   the tangent line    has a

zero    slope,  we  first   take    the derivative:     =   −3  −   2x. Now we  need    to  find    where   the slope   is

zero.   The derivative  −3  −   2x  =   0   at  x   =   − . Now,    we  need    to  find    the y-coordinate,   which   we

get by  plugging    x   =   − into   the     equation    for    y:  8   −   3    .   Therefore,  the

answer  is  .


  1. a = 1, b = 0, and c = 1.


The two equations   will    have    a   common  tangent line    where   they    have    the same    slope,  which   we

find    by  taking  the derivative  of  each    equation.   The derivative  of  the first   equation    is:     =   2x  +

a.  The derivative  of  the second  equation    is      =   c   +   2x. Setting the two derivatives equal   to

each    other,  we  get a   =   c.  Each    equation    will    pass    through the point   (−1,    0). If  we  plug    (−1,    0)

into    the first   equation,   we  get 0   =   (−1)^2  +   a(−1)   +   b,  which   simplifies  to: a   −   b   =   1.  If  we  plug
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