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if the tangent line to y= f(x) at (4, -5) passes through the point (1,1), find f(4) and the equation of the tangent line at 4
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to define any line you need two points. In your problem you got that: 1 point is the point of tngency (4,-5), the other is (1,1). So your line is: (y,x)=(1,1)+c(4-1,-5-1)
how did u arrive at that please? i wanna know it
the equation of the line I wrote you is in solcolled point vector form, where point is (1,1) and direction vector is (4-1,-5-1), and c is arbitrary real constant
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