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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =
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12 Aug 2018, 06:38
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
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
Re: Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =
[#permalink]
12 Aug 2018, 18:09
12 Aug 2018, 18:09
2
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
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 \(f(k-1) = (k - 1)^2 - 5 = k^2 - 2k + 1 - 5 = k^2 -2k - 4\)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Challenge: If f(3x + 2) = 9x² + 12x - 1, then f(k - 1) =
[#permalink]
14 Aug 2018, 05:28
14 Aug 2018, 05:28
3
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
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
Cheers,
Brent
