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If the function f is defined for all x by f(x) = . . .

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Source: — Data Sufficiency |
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by DavidG@VeritasPrep » Fri Nov 10, 2017 6:39 am
Vincen wrote:If the function f is defined for all x by $$f(x)=ax^2+bx−43,$$ where a and b are constants, what is the value of f(3)?

(1) f(4) = 41
(2) 3a + b = 17

The OA is B .

I got confused here. I don't know how to use the function f(x). Experts, can you give me some help?
It's helpful to think of functions in terms of inputs and outputs. For instance, f(3) is what we get when '3', our input, is plugged into some expression

If $$f(x)=ax^2+bx-43$$ then f(3) = $$a3^2+3b-43,$$ = 9a + 3b - 43.

Rephrase: what is the value of 9a + 3b - 43?

Statement 1: If f(4) = 41 then $$f(4)=a4^2+b4-43,$$ = 41, or 16a + 4b -43 = 41. There's no way this can allow us to solve for 9a + 3b - 43. Statement 1 is not sufficient.

Statement 2: Let's see if there's a way that we can substitute 3a + b = 17 into 9a + 3b - 43. Well, 9a + 3b -43 can be written as 3(3a + b) - 43.

Now we can substitute 17 in place of 3a + b, so 3(3a + b) - 43 becomes 3*17 - 43. Clearly, we'll get a unique value here. No need to solve. This is sufficient. The answer is B
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by Matt@VeritasPrep » Fri Nov 10, 2017 1:17 pm
Let's start by plugging 3 into the function:

f(3) = 3²*a + 3*b - 43

f(3) = 9a + 3b - 43

So to solve, we need to know the value of 9a + 3b. We can write this as 3 * (3a + b), so 3a + b would do the trick too.

S1:: not sufficient, doesn't give 3a + b
S2:: sufficient, DOES give the value of 3a + b

If functions are giving you trouble, try working through this great tutorial, it should help a lot! :)
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