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

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

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New post 12 Aug 2018, 07:39
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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

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

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New post 14 Aug 2018, 06:27
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GMATPrepNow 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

Answer: C

Cheers,
Brent
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If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =  [#permalink]

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New post 12 Aug 2018, 07:58
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GMATPrepNow 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, \(f(3x + 2)\) = \(9x^2 + 12x - 1\)=\(9x² + 12x +4-4-1=(3x+2)^2 - 5\)
Or, \(f(k) = k^2 - 5\) , where k=3x+2

So, \(f(k-1)=(k-1)^2-5=k^2-2k+1-5=k^2-2k-4\)

Ans. (C)
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Re: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =  [#permalink]

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New post 12 Aug 2018, 08:01
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GMATPrepNow 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

*Kudos for all correct solutions


3x+2 = k-1
3x = k-3

Put the value of 3x in 9x^2 + 12x - 1

f(k-1)=f(3x+2)= 9x^2 + 12x - 1
f(k-1)=f(3x+2)= 3x*3x + 4*3x -1 now put 3x = k-3
f(k-1)=f(3x+2)= (k-3)(k-3) + 4(k-3) -1
f(k-1)=f(3x+2)= k^2 - 2k - 4
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If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =  [#permalink]

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New post 12 Aug 2018, 08:03
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GMATPrepNow 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

*Kudos for all correct solutions


f(3x + 2) =\(9x^2 + 12x - 1\)
We see that \((3x+2)^2\) = \(9x^2 + 12x + 4\) , if we subtract 5 from this, we get the desired value.
So, f(3x + 2) = \((3x+2)^2 - 5\) = \(9x^2 + 12x - 1\)
f(k-1) = \((k-1)^2-5\) = \(k^2 - 2k +1 - 5\) = \(k^2- 2k - 4\)
Answer C.
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Re: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =  [#permalink]

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New post 12 Aug 2018, 09:38
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3x+2 = k-1 x= k-3/3
9(k-3/3)^2 +12(k-3/3) -1
K^2 + 9 -6k +4k -12 -1
K^2 -2k -4

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

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New post 12 Aug 2018, 10:24
1
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)^2 = 9x^2+12x−1
(3x+2)^2 =9x^2+12x+4
F(x)=x^2-5
So f(k-1)=k² - 2k - 4

B option

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

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New post 12 Aug 2018, 11:23
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PKN wrote:
GMATPrepNow wrote:
If f(3x + 2) = 9x² + 12x - 1[/color], then f(k - 1) =

A) k² - 2k - 6
B) k² - 2k - 5
C) k² - 2k - 4
D) k² - 2k + 1
E) k² - 2k + 5


Alternate approach:

Let k=1, with the result that f(k-1) = f(0).
The question stem becomes:
If f(3x + 2) = 9x² + 12x - 1, what is the value of f(0)?

Since f(3x+2) = f(0) when x=-2/3, plug x=-2/3 into the expression in blue:
\(9(-\frac{2}{3})^2 + 12(-\frac{2}{3}) - 1 = 4 - 8 - 1 = -5\)

Now plug k=1 into the answer choices to see which yields the target value of -5.
Only C works:
\(k^2 - 2k - 4 = 1 - 2 - 4 = -5\)


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

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New post 12 Aug 2018, 12:41
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HKD1710 wrote:
GMATPrepNow 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

*Kudos for all correct solutions


3x+2 = k-1
3x = k-3

Put the value of 3x in 9x^2 + 12x - 1

f(k-1)=f(3x+2)= 9x^2 + 12x - 1
f(k-1)=f(3x+2)= 3x*3x + 4*3x -1 now put 3x = k-3
f(k-1)=f(3x+2)= (k-3)(k-3) + 4(k-3) -1
f(k-1)=f(3x+2)= k^2 - 2k - 4


awesome approach
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Re: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =  [#permalink]

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New post 13 Aug 2018, 01:46
PKN wrote:
GMATPrepNow 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, \(f(3x + 2)\) = \(9x^2 + 12x - 1\)=\(9x² + 12x +4-4-1=(3x+2)^2 - 5\)
Or, \(f(k) = k^2 - 5\) , where k=3x+2

So, \(f(k-1)=(k-1)^2-5=k^2-2k+1-5=k^2-2k-4\)

Ans. (C)


In the any algebraic approach from above, why equating (3x + 2)=k-1?? I do not understand this.

Anyone can help? Thanks in advance
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Re: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =  [#permalink]

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New post 13 Aug 2018, 02:01
1
GMATPrepNow 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

*Kudos for all correct solutions


OA:C
Let \(3x+2 = y\)

\(3x =y-2\)

\(x= \frac{y-2}{3}\)

\(f(3x + 2) = 9x² + 12x - 1\)

\(f(y) = 9 (\frac{y-2}{3})^2 +12(\frac{y-2}{3}) -1\)

\(f(y) = (y-2)^2 +4(y-2) -1=y^2+4-2*2y + 4y -8 -1 =y^2-5\)

\(f(y)=y^2-5\)

\(f(k - 1) =(k-1)^2-5 =k^2+1-2k-5 =k^2-2k-4\)
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Re: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =  [#permalink]

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New post 14 Aug 2018, 08:33
GMATPrepNow wrote:

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

Answer: C

Cheers,
Brent


Dear Brent,

Very enlightening answer. Your way above answered my question in simple terms.

Thanks
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Re: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =  [#permalink]

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New post 14 Aug 2018, 09:08
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Thanks Mo2men!

Cheers
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Re: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =   [#permalink] 14 Aug 2018, 09:08
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