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

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GreenlightTestPrep
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Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) = [#permalink] New post   12 Aug 2018, 06:38
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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
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C
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pranab223
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Re: Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) = [#permalink] New post   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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Re: Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) = [#permalink] New post   14 Aug 2018, 05:28
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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



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) = y² - 5
And f(7) = 7² - 5

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

Answer: C

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Brent
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Re: Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) = [#permalink] New post   16 Aug 2020, 16:47
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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] New post   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

Final Answer: C
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Re: Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) = [#permalink] New post   02 Feb 2023, 10:51
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) = [#permalink]
02 Feb 2023, 10:51
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