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

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Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) = [#permalink]  12 Aug 2018, 06:38
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Question Stats:

72% (02:09) correct 27% (00:47) wrong based on 29 sessions
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

* I'll post a solution in 2 days
[Reveal] Spoiler: OA

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Brent Hanneson – Creator of greenlighttestprep.com
If you enjoy my solutions, you'll like my GRE prep course.

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Re: Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) = [#permalink]  12 Aug 2018, 18:09
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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

* I'll post a solution in 2 days

Here

$$f(3x + 2) = 9x^2 + 12x - 1 = (3x)^2 + 2*(3x) * 2 +2^2 - 5 = (3x + 2)^2 - 5$$(writing in the form of$$(a + b)^2$$)

Now it can be written as $$f(k-1) = (k - 1)^2 - 5 = k^2 - 2k + 1 - 5 = k^2 -2k - 4$$
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GRE Instructor
Joined: 10 Apr 2015
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Kudos [?]: 4513 [1] , given: 69

Re: Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) = [#permalink]  14 Aug 2018, 05:28
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Expert's post
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 + 4
This 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 - 5

In other words, f(something) = (something)² - 5
So, for example, f(y) = - 5
And f(7) = - 5

Likewise, f(k - 1) = (k - 1)² - 5
= k² - 2k + 1 - 5
= k² - 2k - 4

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com
If you enjoy my solutions, you'll like my GRE prep course.

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Re: Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) = [#permalink]  16 Aug 2020, 16:47
Quick way to solve this one:

Let $$x = 0$$. Then $$f(2) = -1$$

Let $$k = 3$$. Then we get $$f(2)$$

So Plug in 3 into the answer choices for $$k$$ and you're answer should give $$-1$$.

That choice is C.
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Re: Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) = [#permalink]  16 Aug 2020, 17:48
f(3x+2)
=9x^2+12x-1
=(3x)^2+4(3x)-1
=([3x+2]-2)^2 + 4([3x+2]-2) - 1

Therefore f in general looks like
f(y)=(y-2)^2 + 4(y-2) -1

Therefore
f(k-1)
=([k-1]-2)^2 + 4([k-1]-2) - 1
=(k-3)^2 + 4(k-3) -1
= k^2-6k+9+4k-12-1
=k^2 - 2k -4

Re: Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =   [#permalink] 16 Aug 2020, 17:48
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