Finding f(3) for a quadratic function with three known points

[HappMatt]

Im doing a practice test for a asmaytic(not sure how its spelled) and i have the answer but am just unsure of how i was supposed to figure it out. Alright, here's the problem.

if "f" is a quadriatic function with f(0)=0, f(1)=2, and f(2)=3, then f(3)=?
a.4 b.3 c.13/4 E.4

the answer is supposed to be b. but i have no clue how they got that.

 

[arildno]

Hint: What does it mean that f is "a quadratic function"?

 

[HappMatt]

well I am not to sure besideds the fact that i think its a parabola

 

[HallsofIvy]

No, a parabola is a graph. It is true that the graph of any quadratic function is a parabola. A quadratic function is a function that can be written in the form f(x)= ax²+ bx+ c- that's what you are thinking of. I suspect that any test asking such a question will assume you know more than that! Do you notice there are 3 "unknown" coefficients in that? Putting in the three values given for x and y will give you three equations to solve for a, b, c. Once you know that, you can find f(3).
There are easier ways to do this problem but that is the most straightforward.

 

[HappMatt]

well I am still working on this, i just got off of work and I am definitely getting a litle annoyed since i still can't figure this out. the reson i brought up the fact that's ist a parabola is that from the best of my knowledge a porabola is never a horizontal line such as y=3 yet, if f(2)=3 and f(3)=3 the what we are seeing is a horizonal line and that should be impossible in a quadriatic for many reasons. I did a system of equations and found an equation that works for all given values yet of course when f(3) was entered into the equation its output was 6. I used the equation(-1/2)x^2+(5/2)x+0. As for the test its actually just for extra credit and is different from what i am currently studying in that the test is more algebra based and at my school you have to be in calc 1 or precalc to take it so right now I am actually doing a lot more with dirivatives than i am with algebra, so this stuff isn't exactly the freshish in my head.

 

[HallsofIvy]

well I am still working on this, i just got off of work and I am definitely getting a litle annoyed since i still can't figure this out. the reson i brought up the fact that's ist a parabola is that from the best of my knowledge a porabola is never a horizontal line such as y=3 yet, if f(2)=3 and f(3)=3 the what we are seeing is a horizonal line and that should be impossible in a quadriatic for many reasons. I did a system of equations and found an equation that works for all given values yet of course when f(3) was entered into the equation its output was 6. I used the equation(-1/2)x^2+(5/2)x+0. As for the test its actually just for extra credit and is different from what i am currently studying in that the test is more algebra based and at my school you have to be in calc 1 or precalc to take it so right now I am actually doing a lot more with dirivatives than i am with algebra, so this stuff isn't exactly the freshish in my head.

f(x)= (-1/2)x^2+(5/2)x is, in fact, the correct formula.
f(0)= 0, f(1)= -1/2+ 5/2= 2, f(2)= (-1/2)(4)+ (5/2)(2)= -2+ 5= 3 and
f(3)= (-1/2)(9)+(5/2)(3)= (-9+ 15)/2= 6/2= 3.
Apparently, you simply forgot the "2" in the denominator!