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 + 4This 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 - 5In other words, f(
something) =
(something)² - 5So, for example, f(
y) =
y² - 5And f(
7) =
7² - 5Likewise, f(
k - 1) =
(k - 1)² - 5=
k² - 2k + 1 - 5= k² - 2k - 4
Answer: C
Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep