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Question Number 130713 by harckinwunmy last updated on 28/Jan/21

Commented by EDWIN88 last updated on 28/Jan/21

ans: B

$${ans}:\:{B} \\ $$

Answered by EDWIN88 last updated on 28/Jan/21

g(x)=a(x+2)(x−3)(x−6)  g(0)=a(2)(−3)(−6)<0               36a<0; a<0  (1)g(−3)=a(−1)(−6)(−9)>0  (2)g(−1)=a(1)(−4)(−7)<0  (3)g(4)=a(6)(1)(−2)>0  (4)g(5)=a(7)(2)(−1)>0

$${g}\left({x}\right)={a}\left({x}+\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{6}\right) \\ $$$${g}\left(\mathrm{0}\right)={a}\left(\mathrm{2}\right)\left(−\mathrm{3}\right)\left(−\mathrm{6}\right)<\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{36}{a}<\mathrm{0};\:{a}<\mathrm{0} \\ $$$$\left(\mathrm{1}\right){g}\left(−\mathrm{3}\right)={a}\left(−\mathrm{1}\right)\left(−\mathrm{6}\right)\left(−\mathrm{9}\right)>\mathrm{0} \\ $$$$\left(\mathrm{2}\right){g}\left(−\mathrm{1}\right)={a}\left(\mathrm{1}\right)\left(−\mathrm{4}\right)\left(−\mathrm{7}\right)<\mathrm{0} \\ $$$$\left(\mathrm{3}\right){g}\left(\mathrm{4}\right)={a}\left(\mathrm{6}\right)\left(\mathrm{1}\right)\left(−\mathrm{2}\right)>\mathrm{0} \\ $$$$\left(\mathrm{4}\right){g}\left(\mathrm{5}\right)={a}\left(\mathrm{7}\right)\left(\mathrm{2}\right)\left(−\mathrm{1}\right)>\mathrm{0} \\ $$

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