Working with functions

Question

Suppose $f(3-x)=2x^2-5x+4$ and $f(x)=ax^2+bx+c$. What is $a+b+c$?

I don't know how to approach this. I thought of maybe doing $a(3-x)^2+b(3-x)+c=2x^2-5x+4$ and solving for $a+b+c$ but it got messy.

We have, $$a+b+c=f(1)=f(3-2)=2(2^2)-5(2)+4=2.$$ – Juniven Acapulco Apr 07 '17 at 02:10 — Juniven Acapulco, Apr 07 '17 at 02:10
@ΘΣΦGenSan You should write that up as an answer. – amd Apr 07 '17 at 02:15 — amd, Apr 07 '17 at 02:15

score 3 · Accepted Answer · answered Apr 07 '17 at 02:19

3

We know that $f(x)=ax^2+bx+c$ and so $f(1)=a+b+c$. Also, $f(3-x)=2x^2-5x+4$ and hence we get $$a+b+c=f(1)=f(3−2)=2(2^2)−5(2)+4=2.$$

answered Apr 07 '17 at 02:19

Juniven Acapulco

score 1 · Answer 2 · answered Apr 07 '17 at 02:06

On continue from your answer.

$a(9+x^2-6x)+b(3-x)+c=2x^2-5x+4$

$\implies 9a+9ax^2-6ax+3b-3bx+c=2x^2-5x+4$

$\implies 9ax^2-(6a+3b)x+(9a+3b+c)=2x^2-5x+4$

Comparing both sides,

$$9a=2$$

$$6a+3b=5$$

$$9a+3b+c=4$$

Find a from first equation, then put a in second equation to find b.

Then put a, b in third equation to find c.

2 Answers2