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Consider the curve defined by -8x^2 +5xy+y^3=149. Write an equation for the line tangent to the curve at the point (4, -1)
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slope of line tangent to a curve is got by differentiating the curve
and substitute (4,-1) in it
then use slope pont form
would you differentiate the curve with dy/dx?
why dont u try that yourself,,it isnt a hard task ...i will check ur answer
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I got dy/dx= (16x+5y)/(5x+3y)
it is dy/dx=16x-5y/(5x+3y^2)
now substitute (4,-1) to get slope at that point
oh ok i see where i messed up
equation of tangent will be y+1=m(x-4) where m is slope at the point
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got it?
i got 3 for the slope and i think i understand the rest thank you
you are welcome
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