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