# $f(x - 1) + f(x − 2)$ and the sum of coeficients

If $f(x-1)+f(x-2) = 5x^2 - 2x + 9$

and

$f(x)= ax^2 + bx + c$

what would be the value of $a+b+c$?

I was doing

$f(x-1)+f(x-2)= f(x-3)$ then $f(x)$

a = 5
b = -2
c = 9


$(5-3)+(-2-3)+(9-3)$

But do not think is is correct

What would be correct approach?

$$f(x-1)+f(x-2) = 5x^2 - 2x + 9$$and $$f(x)=ax^2+bx+c$$ for $x=1,2,3$ we get the system $$f(0)+f(-1)=a-b+2c =12$$ $$f(1)+f(0)=a+b+2c=25$$ $$f(2)+f(1)=5a+3b+2c = 48$$ with solutions $$a=5/2,b=13/2,c=8$$ so $$f(x)=\frac{5}{2}x^2+\frac{13}{2}x+8$$

If $f(x) = ax^2 + bx+ c$, what is $f(x-1)$ and what is $f(x-2)$?

Work those out, then add both expressions to equate to $5x^2-2x+9$.

$$f''(x)=2a, 4a=f''(x-1)+f''(x-2)=10, a=2.5$$

$$f'(x)=2ax+b=5x+b, 5(x-1)+5(x-2)+2b$$

$$f'(x-1)+f'(x-2)=10x-2, -15+2b=-2, b=6.5$$

$$f(x)=ax^2+bx+c, 2.5(x-1)^2+6.5(x-1)+2.5(x-2)^2+6.5(x-2)+2c=f(x-1)+f(x-2)=5x^2-2x+9$$

let $x=0$ on both side, $2.5-6.5+10-13+2c=9, c=8$

so $a+b+c=2.5+6.5+8=17$

We know $$f(x-1)=a(x-1)^2+b(x-1)+c\quad\text{and}\quad f(x-2)=a(x-2)^2+b(x-2)+c.$$ Expand these terms, add them, and combine like terms via powers of $x$. Now you can get three equations in three variables by equating the coefficients of the left with the right since you know $$f(x-1)+f(x-2)=5x^2-2x+9.$$

• So I have done: $f(x-1)= a(x-1)^2 + b(x-1) + c = a(x-1)(x-1) + bx - b + c = ax^2 -2ax + a + bx - b + c$ $f(x-2) = a(x-2)^2 + b(x-2) + c = a(x-2)(x-2) + bx - 2b + c = ax^2 -4ax + 4 + bx - b + c$ $so $$f(x−1)+f(x−2) = 2ax^2 -6ax + a +4 + 2bx - 2b + 2c = 5x^2 - 2x + 9$$ right? – edgarmtze Mar 28 '13 at 20:47 • @cMinor check again your expansion of$f(x-2)\$. It should be 4a and -2b. – Jerry Mar 28 '13 at 20:52