# f(x)=ax^2-5b For the function f defined above, a and b are constants and f(x+2)=5. What is the value of f(-x-2)? a) -5 b) 0 c) 5 d) 10?

The answer is not A

### 6 Answers

- PhilipLv 61 month ago
f(x) = ax^2 - 5b. f(x+2) = 5, ie., a(x+2)^2 -5b = 5;

f(-x-2)= f(-(x+2)) = a(-(x+2))^2 -5b = a(x+2)^2 -5b = f(x+2) = 5. Answer = c).

- KrishnamurthyLv 71 month ago
f(x) = ax^2 - 5b

For the function f defined above, a and b are constants

and f(x + 2) = 5.

The value of f(-x - 2):

a) -5

- husoskiLv 71 month ago
There's really no reason to expect (a) to be the answer. The only use of x in f(x) is to square it, and there's no simple way to tweak x to make the sign of f(x) flip, considering that -5b is going to be added.

What I'd guess you're supposed to see is that the f(x) is a polynomial with only even powers of x, which means that f(-x) = f(x) for all x. Such a function sometimes called "even", which is short for "symmetrical with respect to the x=0 line".

Then, I'll bet you're supposed to recognize that -x-2 = -(x + 2). Put those together and:

f(-x - 2) = f(-[x+2]) . . . simplifying the argument to f

= f(x + 2) . . . because f is "even"

= 5 . . . because f(x+2) = 5 was given

Write these out until you are comfortable with the pattern. Soon, you'll be able to answer a question like this with essentially no calculation at all, just by recognizing symmetry.

If a polynomial f(x) has only odd powers of x instead, then you'd have f(-x) = -f(x). Such a function is called "odd", short for "antisymmetrical with respect to the x=0 line". In that case, if f(x + 2) = 5 were true, then f(-x - 2) would be -5. Again, almost no calculation is needed...just a few sign flips.

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- Anonymous1 month ago
Sorry, my answer is "Anonymous" as your post is Anonymous. There is no need for an Anonymous question in this category, Most 'Anonymous' posts are trolls, maybe yours is not, but the high likelihood is there. If you post as yourself, I would be glad to help you with any problem you have.

I spend a lot of time on each problem, I just want to make sure it's for a worthwhile cause.