Menu Close

if-f-x-ax-2-bx-c-and-f-x-3-7x-2-2x-5-find-1-a-b-c-2-a-b-c-

Tinku Tara 0 Comments
Pin




Question Number 151081 by malwan last updated on 18/Aug/21
if f(x)=ax^2 +bx+c and f(x+3)=7x^2 +2x+5 find 1) a+b+c 2) a−b+c
$${if}\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${and}\:{f}\left({x}+\mathrm{3}\right)=\mathrm{7}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\:{find} \\ $$$$\left.\mathrm{1}\left.\right)\:{a}+{b}+{c}\:\:\:\:\:\mathrm{2}\right)\:{a}−{b}+{c} \\ $$
Answered by mr W last updated on 18/Aug/21
f(1)=a+b+c=f(−2+3)=7(−2)^2 +2(−2)+5=29 f(−1)=a−b+c=f(−4+3)=7(−4)^2 +2(−4)+5=109
$${f}\left(\mathrm{1}\right)={a}+{b}+{c}={f}\left(−\mathrm{2}+\mathrm{3}\right)=\mathrm{7}\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}\left(−\mathrm{2}\right)+\mathrm{5}=\mathrm{29} \\ $$$${f}\left(−\mathrm{1}\right)={a}−{b}+{c}={f}\left(−\mathrm{4}+\mathrm{3}\right)=\mathrm{7}\left(−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{2}\left(−\mathrm{4}\right)+\mathrm{5}=\mathrm{109} \\ $$
Commented by malwan last updated on 18/Aug/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 18/Aug/21
NiCe!
$$\mathbb{N}{i}\mathbb{C}{e}! \\ $$
Answered by Rasheed.Sindhi last updated on 18/Aug/21
f(x)=ax^2 +bx+c ∧ f(x+3)=7x^2 +2x+5 1)a+b+c=? 2)a−b+c=? −.−.−.−.−.−.−.− a(x+3)^2 +b(x+3)+c=7x^2 +2x+5 a(x^2 +6x+9)+b(x+3)+c=7x^2 +2x+5 ax^2 +(6a+b)x+9a+3b+c=7x^2 +2x+5 a=7 ∧ 6a+b=2 ∧ 9a+3b+c=5 a=7⇒b=−40⇒c=5−63+120=62 a+b+c=7+(−40)+62=29 a−b+c=7−(−40)+62=109
$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}\:\wedge\:{f}\left({x}+\mathrm{3}\right)=\mathrm{7}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5} \\ $$$$\left.\mathrm{1}\left.\right){a}+{b}+{c}=?\:\:\:\:\mathrm{2}\right){a}−{b}+{c}=? \\ $$$$−.−.−.−.−.−.−.− \\ $$$${a}\left({x}+\mathrm{3}\right)^{\mathrm{2}} +{b}\left({x}+\mathrm{3}\right)+{c}=\mathrm{7}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5} \\ $$$${a}\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}\right)+{b}\left({x}+\mathrm{3}\right)+{c}=\mathrm{7}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5} \\ $$$${ax}^{\mathrm{2}} +\left(\mathrm{6}{a}+{b}\right){x}+\mathrm{9}{a}+\mathrm{3}{b}+{c}=\mathrm{7}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5} \\ $$$${a}=\mathrm{7}\:\wedge\:\mathrm{6}{a}+{b}=\mathrm{2}\:\wedge\:\mathrm{9}{a}+\mathrm{3}{b}+{c}=\mathrm{5} \\ $$$${a}=\mathrm{7}\Rightarrow{b}=−\mathrm{40}\Rightarrow{c}=\mathrm{5}−\mathrm{63}+\mathrm{120}=\mathrm{62} \\ $$$${a}+{b}+{c}=\mathrm{7}+\left(−\mathrm{40}\right)+\mathrm{62}=\mathrm{29} \\ $$$${a}−{b}+{c}=\mathrm{7}−\left(−\mathrm{40}\right)+\mathrm{62}=\mathrm{109}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by malwan last updated on 18/Aug/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 18/Aug/21
f(x)=ax^2 +bx+c_((i)) ∧ f(x+3)=7x^2 +2x+5_((ii)) 1)a+b+c 2)a−b+c ⇚≪−.−.−.−≫⇛ (i): f(x)=ax^2 +bx+c..........A (ii)⇒f(x−3+3)=7(x−3)^2 +2(x−3)+5 ⇒f(x)=7x^2 −40x+62.....B A & B⇒a=7,b=−40,c=62 a+b+c=7−40+62=29 a−b+c=7+40+62=109
$$\underset{\left({i}\right)} {\underbrace{{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}}}\:\wedge\:\underset{\left({ii}\right)} {\underbrace{{f}\left({x}+\mathrm{3}\right)=\mathrm{7}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}} \\ $$$$\left.\mathrm{1}\left.\right){a}+{b}+{c}\:\:\:\:\mathrm{2}\right){a}−{b}+{c} \\ $$$$\Lleftarrow\ll−.−.−.−\gg\Rrightarrow \\ $$$$\left({i}\right):\:\:\:\:\:\:\:\:\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}……….{A} \\ $$$$\left({ii}\right)\Rightarrow{f}\left({x}−\mathrm{3}+\mathrm{3}\right)=\mathrm{7}\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{2}\left({x}−\mathrm{3}\right)+\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow{f}\left({x}\right)=\mathrm{7}{x}^{\mathrm{2}} −\mathrm{40}{x}+\mathrm{62}…..{B} \\ $$$${A}\:\&\:{B}\Rightarrow{a}=\mathrm{7},{b}=−\mathrm{40},{c}=\mathrm{62} \\ $$$${a}+{b}+{c}=\mathrm{7}−\mathrm{40}+\mathrm{62}=\mathrm{29} \\ $$$${a}−{b}+{c}=\mathrm{7}+\mathrm{40}+\mathrm{62}=\mathrm{109} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *