Question #04999

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Answer 1 · 2016-10-15T08:46:22

#p > 1/2 (6 + sqrt[36 + f^2])#

If #y >0 forall x# then #p > 0# because #p# is the coeficient of #x^2# and #y(x)# cannot have #x# crossings so #y(0) = p-6 > 0#

Anyway the #x# crossings are given at

#x=(-f pm sqrt[f^2 + 24 p - 4 p^2])/(2 p)# so if

#f^2+24p-4p^2<0# no feasible crossings

or solving for #p#

#p < 1/2 (6 - sqrt[6^2 + f^2]) # and #p > 1/2 (6 + sqrt[6^2 + f^2])#

Concluding

#p > 6# and #p > 1/2 (6 + sqrt[6^2 + f^2])# or finally

#p > 1/2 (6 + sqrt[6^2 + f^2])#