SOLUTION: Which of the following is another version of {{{f(x) = 2x^2 - 4x - 7}}} ? a. {{{f(x) = 2(x -1)^2 - 9}}} b. {{{f(x) = (x - 1)^2 - 9}}} c. {{{f(x) = (x + 1)^2 - 1/2}}} d. {{{f(x

Algebra ->  Equations -> SOLUTION: Which of the following is another version of {{{f(x) = 2x^2 - 4x - 7}}} ? a. {{{f(x) = 2(x -1)^2 - 9}}} b. {{{f(x) = (x - 1)^2 - 9}}} c. {{{f(x) = (x + 1)^2 - 1/2}}} d. {{{f(x      Log On


   



Question 141506: Which of the following is another version of f%28x%29+=+2x%5E2+-+4x+-+7 ?
a. f%28x%29+=+2%28x+-1%29%5E2+-+9
b. f%28x%29+=+%28x+-+1%29%5E2+-+9
c. f%28x%29+=+%28x+%2B+1%29%5E2+-+1%2F2
d. f%28x%29 cannot be factored; prime

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Which of the following is another version of f%28x%29+=+2x%5E2+-+4x+-+7 ?
a. f%28x%29+=+2%28x+-1%29%5E2+-+9
b. f%28x%29+=+%28x+-+1%29%5E2+-+9
c. f%28x%29+=+%28x+%2B+1%29%5E2+-+1%2F2
d. f%28x%29 cannot be factored; prime

We can eliminate d because none of the choices are
factorizations of f(x) anyway.

The easy way to find which one is correct is to substitute
an arbitrary value for x, (0 is a good choice since it's easy)
in the original and in each of the choices.

Substituting x=0 in the original:

f%28x%29+=+2x%5E2+-+4x+-+7
f%280%29+=+2%280%29%5E2+-+4%280%29+-+7
f%280%29+=+-7

Now substitute x=0 in the answer a.

f%28x%29+=+2%28x-1%29%5E2+-+9
f%280%29+=+2%280-1%29%5E2+-+9
f%280%29+=+2%28-1%29%5E2+-+9 
f%28x%29+=+2%281%29+-+9
f%28x%29+=+2+-+9
f%28x%29+=+-7

So a. is possible.

Now substitute x=0 in the answer b.

f%28x%29+=+%28x-1%29%5E2+-+9
f%280%29+=+%280-1%29%5E2+-+9
f%280%29+=+%28-1%29%5E2+-+9 
f%28x%29+=+%281%29+-+9
f%28x%29+=++1-+9
f%28x%29=-8

So b. is NOT possible.

Now substitute x=0 in the choice c.

f%28x%29+=+%28x+%2B+1%29%5E2+-+1%2F2
f%28x%29+=+%28x%2B1%29%5E2+-+1%2F2
f%280%29+=+%280%2B1%29%5E2+-+1%2F2
f%280%29+=+%281%29%5E2+-+1%2F2 
f%28x%29+=+2%281%29+-+1%2F2
f%28x%29+=+2+-+1%2F2
f%28x%29+=+4%2F2-1%2F2
f%28x%29+=+3%2F2

So c. is NOT possible.

Thus the only possible choice is a.

--------------

However, that's not what your teacher wants you
to do, although it is one way to get the correct
answer on a multiple choice test. I just thought 
I'd show you the above method of substituting in
each of the answers to rule out the incorrect 
choices. This may save you from having to miss a 
problem which you can't solve on a multiple 
choice test. Of course, this method will not work 
on the kind of test where you have to show your 
work.

-------------

Here is what your teacher wanted you to do:

f%28x%29+=+2x%5E2+-+4x+-+7

Factor 2 out of the first two terms only:

f%28x%29+=+2%28x%5E2-2x%29+-+7

We will complete the square inside the parentheses:

Multiply the coefficient of x, which is -2, by 1%2F2,
getting -1.  Then square -1 getting +1

So add +1 and immediately subtract 1 inside the 
parentheses:

f%28x%29+=+2%28x%5E2-2x%2B1-1%29+-+7

Change the parentheses to brackets so they can hold 
parentheses:

f%28x%29+=+2[x%5E2-2x%2B1-1]+-+7

Factor only the first three terms inside the brackets:

f%28x%29+=+2[%28x-1%29%28x-1%29-1]+-+7

f%28x%29+=+2[%28x-1%29%5E2-1]+-+7

Now remove the brackets by distributing the 2, leaving
the %28x-1%29%5E2 intact:

f%28x%29+=+2%28x-1%29%5E2-2+-+7

Combine the -2 and the -7 and get -9

f%28x%29+=+2%28x-1%29%5E2-9

which is choice a.

Edwin