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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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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

*Kudos for all correct solutions

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

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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

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

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12 Aug 2018, 07:58
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

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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12 Aug 2018, 08:01
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

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

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12 Aug 2018, 09:38
1
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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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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12 Aug 2018, 11:23
1
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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12 Aug 2018, 12:41
1
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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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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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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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

Cheers,
Brent

Dear Brent,

Thanks
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Joined: 12 Sep 2015
Posts: 3847
Re: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =  [#permalink]

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14 Aug 2018, 09:08
Top Contributor
Thanks Mo2men!

Cheers
Brent

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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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