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Question-123922




Question Number 123922 by help last updated on 29/Nov/20
Answered by liberty last updated on 29/Nov/20
y=p(x−3)^2 +1 = px^2 −6px+9p+1   { ((p=−a)),((b=−6p )),((c=9p+1)) :}
$${y}={p}\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{1}\:=\:{px}^{\mathrm{2}} −\mathrm{6}{px}+\mathrm{9}{p}+\mathrm{1} \\ $$$$\begin{cases}{{p}=−{a}}\\{{b}=−\mathrm{6}{p}\:}\\{{c}=\mathrm{9}{p}+\mathrm{1}}\end{cases} \\ $$
Commented by help last updated on 29/Nov/20
value of p?
$${value}\:{of}\:{p}? \\ $$
Commented by help last updated on 29/Nov/20
from here we can say option A  is valid...i solve with graph tho  thanks
$${from}\:{here}\:{we}\:{can}\:{say}\:{option}\:{A} \\ $$$${is}\:{valid}…{i}\:{solve}\:{with}\:{graph}\:{tho} \\ $$$${thanks} \\ $$

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