GreenlightTestPrep wrote:

If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =

A) k² - 2k - 6

B) k² - 2k - 5

C) k² - 2k - 4

D) k² - 2k + 1

E) k² - 2k + 5

Given: f(3x + 2) = 9x² + 12x - 1

First notice that 9x² + 12x - 1

looks a lot like how (3x + 2)² looks when we expand an simplify it.

Notice that

(3x + 2)² = 9x² + 12x + 4This is VERY similar to 9x² + 12x - 1

In fact, if we take

9x² + 12x + 4 and

subtract 5, we get 9x² + 12x - 1

That is:

9x² + 12x + 4 - 5 = 9x² + 12x - 1

So, we can write: f(

3x + 2) = 9x² + 12x - 1

=

9x² + 12x + 4 - 5=

(3x + 2)² - 5In other words, f(

something) =

(something)² - 5So, for example, f(

y) =

y² - 5And f(

7) =

7² - 5Likewise, f(

k - 1) =

(k - 1)² - 5=

k² - 2k + 1 - 5= k² - 2k - 4

Answer: C

Cheers,

Brent

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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