Question Number 12558 by @ANTARES_VY last updated on 25/Apr/17
$$\boldsymbol{\mathrm{This}}\:\: \\ $$$$\boldsymbol{\mathrm{y}}=−\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{6}\boldsymbol{\mathrm{x}}−\mathrm{12}\:\:\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{values}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{function}}\:\boldsymbol{\mathrm{area}}. \\ $$
Answered by ajfour last updated on 25/Apr/17
![y=−12−[(x−3)^2 −9] = −3−(x−3)^2 range of function is y∈ (−∞,−3) .](Q12560.png)
$${y}=−\mathrm{12}−\left[\left({x}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{9}\right] \\ $$$$\:\:=\:−\mathrm{3}−\left({x}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$${range}\:{of}\:{function}\:{is} \\ $$$${y}\in\:\left(−\infty,−\mathrm{3}\right)\:. \\ $$
Commented by mrW1 last updated on 25/Apr/17
![y∈ (−∞,−3]](Q12564.png)
$${y}\in\:\left(−\infty,−\mathrm{3}\right]\: \\ $$$$ \\ $$
Commented by ajfour last updated on 25/Apr/17
$${yes}\:{of}\:{course}. \\ $$