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Question-197967




Question Number 197967 by Blackpanther last updated on 06/Oct/23
Answered by mr W last updated on 06/Oct/23
f(x)=a(x+1)(x−10)  ((20)/3)=a(0+1)(0−10)  ⇒a=−(2/3)  at x=((−1+10)/2)=(9/2):  f((9/2))=−(2/3)((9/2)+1)((9/2)−10)=((121)/6)  eqn. of line:  −(x/1)+(y/b)=1  −(9/2)+((121)/(6b))=1  ⇒b=((11)/3) ←y−intersection of line
$${f}\left({x}\right)={a}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{10}\right) \\ $$$$\frac{\mathrm{20}}{\mathrm{3}}={a}\left(\mathrm{0}+\mathrm{1}\right)\left(\mathrm{0}−\mathrm{10}\right) \\ $$$$\Rightarrow{a}=−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${at}\:{x}=\frac{−\mathrm{1}+\mathrm{10}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{2}}: \\ $$$${f}\left(\frac{\mathrm{9}}{\mathrm{2}}\right)=−\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{9}}{\mathrm{2}}+\mathrm{1}\right)\left(\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{10}\right)=\frac{\mathrm{121}}{\mathrm{6}} \\ $$$${eqn}.\:{of}\:{line}: \\ $$$$−\frac{{x}}{\mathrm{1}}+\frac{{y}}{{b}}=\mathrm{1} \\ $$$$−\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{121}}{\mathrm{6}{b}}=\mathrm{1} \\ $$$$\Rightarrow{b}=\frac{\mathrm{11}}{\mathrm{3}}\:\leftarrow{y}−{intersection}\:{of}\:{line} \\ $$
Commented by mr W last updated on 06/Oct/23
Commented by Blackpanther last updated on 06/Oct/23
thanks

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