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OpenStudy (anonymous):

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)

OpenStudy (aravindg):

slope of line tangent to a curve is got by differentiating the curve

OpenStudy (aravindg):

and substitute (4,-1) in it

OpenStudy (aravindg):

then use slope pont form

OpenStudy (anonymous):

would you differentiate the curve with dy/dx?

OpenStudy (aravindg):

why dont u try that yourself,,it isnt a hard task ...i will check ur answer

OpenStudy (anonymous):

I got dy/dx= (16x+5y)/(5x+3y)

OpenStudy (aravindg):

it is dy/dx=16x-5y/(5x+3y^2)

OpenStudy (aravindg):

now substitute (4,-1) to get slope at that point

OpenStudy (anonymous):

oh ok i see where i messed up

OpenStudy (aravindg):

equation of tangent will be y+1=m(x-4) where m is slope at the point

OpenStudy (aravindg):

got it?

OpenStudy (anonymous):

i got 3 for the slope and i think i understand the rest thank you

OpenStudy (aravindg):

you are welcome

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