Question Number 12596 by @ANTARES_VY last updated on 26/Apr/17
$$\boldsymbol{\mathrm{y}}=−\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{bx}}+\boldsymbol{\mathrm{c}}\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=−\mathrm{1} \\ $$$$\boldsymbol{\mathrm{point}}\:\:\boldsymbol{\mathrm{function}}\:\:\boldsymbol{\mathrm{accepts}}\:\:\:\boldsymbol{\mathrm{a}}\:\:\boldsymbol{\mathrm{maximum}} \\ $$$$\boldsymbol{\mathrm{value}}\:\:\boldsymbol{\mathrm{equal}}\:\:\boldsymbol{\mathrm{to}}\:\:\mathrm{5}.\:\:\boldsymbol{\mathrm{Find}}\:\:\boldsymbol{\mathrm{y}}\left(\mathrm{1}\right). \\ $$
Answered by ajfour last updated on 26/Apr/17
![y=c−[(x−(b/2))^2 −(b^2 /4)] has maximum at x=(b/2) so (b/2)=−1 ⇒ b=−2 the maximum value being c+(b^2 /4) =5 ⇒ c=5−(b^2 /4)= 5−(4/4)=4 y= −x^2 −2x+4 =5−(x+1)^2 y(1)=−1+b+c = −1−2+4=1 .](Q12597.png)
$${y}={c}−\left[\left({x}−\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}\right] \\ $$$${has}\:{maximum}\:{at}\:{x}=\frac{{b}}{\mathrm{2}} \\ $$$${so}\:\frac{{b}}{\mathrm{2}}=−\mathrm{1}\:\:\Rightarrow\:{b}=−\mathrm{2}\: \\ $$$${the}\:{maximum}\:{value}\:{being} \\ $$$$\:\:{c}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}}\:=\mathrm{5}\:\Rightarrow\:{c}=\mathrm{5}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}}=\:\mathrm{5}−\frac{\mathrm{4}}{\mathrm{4}}=\mathrm{4} \\ $$$${y}=\:−{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}\:=\mathrm{5}−\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${y}\left(\mathrm{1}\right)=−\mathrm{1}+{b}+{c}\:=\:−\mathrm{1}−\mathrm{2}+\mathrm{4}=\mathrm{1}\:. \\ $$
Commented by @ANTARES_VY last updated on 26/Apr/17
$$\boldsymbol{\mathrm{error}}\:\:\boldsymbol{\mathrm{answer}} \\ $$
Commented by ajfour last updated on 26/Apr/17
$${now}\:? \\ $$
Commented by @ANTARES_VY last updated on 26/Apr/17
$$\boldsymbol{\mathrm{The}}\:\:\boldsymbol{\mathrm{answer}}\:\:\boldsymbol{\mathrm{should}}\:\:\boldsymbol{\mathrm{be}}\:\:\mathrm{1}. \\ $$
Commented by @ANTARES_VY last updated on 26/Apr/17
$$\boldsymbol{\mathrm{A}}\:\:\boldsymbol{\mathrm{series}}\:\:\boldsymbol{\mathrm{please}} \\ $$