Cracking The SAT Premium

replace the f(x)    portion of  the equation,   so  you can just    write   out the rest    of  it, ,   next    to

each    to  see if  it  can be  true.   Start   with    (A) since   you are looking for the smallest    value   of  f(x).   If

0   =    ,  then    0   =   3x  −   2   when    you square  both    sides.  Add 2   to  both    sides   to  get 2   =   3x, and

then    divide  both    sides   by  3.  You get x   =    .  Since   this    is  a   real    value,  the equation    works,  so  the

smallest    value   of  f(x)    is  0.  Choice  (A) is  correct.

10. 
    

    B Plug in the values from the chart! Use the pair (–3, –7) from the top of the chart and eliminate
    answers that are not true: Choice (A) becomes –7 = –3 – 4, which is true. Keep it. Keep (B): –
    7 = 2(–3) – 1 is true. Get rid of (C), which becomes –7 = 2(–3) + 2: –7 does not equal –4. Get
    rid of (D): –7 = 3(–3) – 3. –7 does not equal –12. Now use another pair just to test (A) and (B).
    Using (–1, –3), (A) gives –3 = –1 – 4, which is not true, so eliminate it, leaving only (B). The
    values (–1, –3) work: –3 = 2(–1) – 1.

    
    


11. 
    

    B First, find the slope of line l by using the slope formula: . A line

    
    

perpendicular   to  line    l   must    have    a   slope   that    is  the negative    reciprocal  of  l’s slope.  So, its

slope   should  be  - . In  the standard    form    of  a   line    Ax  +   By  =   C,  the slope   is   .  Only    (B) has a

slope   of  - . If  you didn’t  remember    the rule    about   the slope   of  perpendicular   lines   in  standard

form,   you could   have    converted   the answers to  slope-intercept form    and sketched    out each    of

the lines   to  look    for the answer  that    looked  perpendicular   to  l.

18. 
    

    B To find the value of f(3) + f(5), find the values of f(3) and f(5) separately: f(3) = 2(3)^2 + 4 = 22
    and f(5) = 2(5)^2 + 4 = 54. So f(3) + f(5) = 76. You can tell that f(4) will be between 22 and 54,
    so you can cross out (A). If you Ballpark (C) and (D), putting 10 or 15 in the function will give
    you a number bigger than 100, and you’re looking for 76, so (C) and (D) are too big. That
    means the answer is (B) by POE.

    
    


19. 
    

    C Remember your transformation rules. Whenever a parabola faces down, the quadratic equation

    
    

has a   negative    sign    in  front   of  x^2     term.   It  always  helps   to  plug    in! Let’s   take    an  example.    If  your

original    equation    was (x  −   2)^2 ,  putting a   negative    sign    in  front   would   make    the parabola    open

downward,   so  you’ll  have    –(x −   2)^2 .  If  you expand  it  out,    you get –x^2    +   4x  −   4.  Notice  that    the

value   of  a   in  this    equation    is  –1. Also    notice  that    the value   of  c   is  –4. This    allows  you to

eliminate   (B) and (D).    Now you must    plug    in  differently to  distinguish between (A) and (C).    Be