Need Help With Function: f(x) = ax2 + bx + c

[Heather]

I have no idea how to begin to solve this problem! I hope somebody can help me!!!
Here's the problem...

Let f(x) = ax2 + bx + c be a quadratic function, where a, b and c are parameters.

(a) Find all real values of a, b and c such that f(x) - f(x - 2) = x.

(b) Given the specific values of a and b in (a) and c = -1, find the zero(s) of the function and the coordinate of the point where the function achieves its minimum or maximum.

 

[stapel]

a) Plug in, and see where it leads:

. . . . .\(\displaystyle \L f(x)\, =\,ax^2\,+\,bx\,+\,c\)

. . . . .\(\displaystyle \L f(x\,-\,2)\,=\,a(x\,-\,2)^2\,+\,b(x\,-\,2)\, +\,c\, =\, ax^2\,+\, (b\,-\,4a)x\,+\,4a\,-\,2b\,+\,c\)

Subtract f(x - 2) from f(x), simplify, and set equal to the specified value. See what relationships you can find.

b) Plug the given values in for the specified variables, and find the zeroes and vertex of the resulting quadratic.

If you get stuck, please reply showing all of your work and reasoning. Thank you!

Eliz.

 

[Heather]

okay I did the first part-plugging in (x-2) but i'm not sure where i went wrong in multiplying it all out because i didn't get the same answer as you...

f(x-2)=a(x-2)^2+b(x-2)+c=ax^2-4ax+4a+bx-2b+c

 

[stapel]

i didn't get the same answer as you...

I typoed the "a" on the "4" in the constant term. Sorry!

What did you get when you subtracted f(x - 2) from f(x)? How much further did you get in the solution?

Eliz.

 

[Heather]

well i subtracted the 2...

f(x)-f(x-2)=ax^2+bx+c-ax^2-4ax+4a+bx-2b+c

=4ax-4a+2b

and now am i supposed to solve for a and b?