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prince50
@prince50
July 2022
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написать уравнение касательной графика функции f(x) = 2x²+4x-10 в точке х0=2
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sedinalana
Verified answer
Уравнение касательной y=f(x0)+f`(x0)*(x-x0)
f(2)=2*4+4*2-10=8+8-10=6
f`(x)=4x+4
f`(2)=4*2+4=8+4=12
y=6+12*(x-2)=6+12x-24=12x-18
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Answers & Comments
Verified answer
Уравнение касательной y=f(x0)+f`(x0)*(x-x0)f(2)=2*4+4*2-10=8+8-10=6
f`(x)=4x+4
f`(2)=4*2+4=8+4=12
y=6+12*(x-2)=6+12x-24=12x-18