Question

Consider the function:(-2x^2 +3x -7) y= -2x^2x +3x +17

Find y ′ using implicit differentiation. Do not solve for y.


What is the slope of the tangent at (x,y) = (−1,−1)?


Find y ′ by solving for y and using the quotient rule.


What is the slope of the tangent when x = -1 ?

Thanks for your help.

Answer 1

(-2x^2 +3x -7) y= -2x^2 +3x +17
(-4x+3)y + (-2x^2+3x-7)y' = -4x+3

y' = (4x-3)(y-1)/(2x^2-3x+7)
y'(-1,-1) = (-7)(-2)/(2+3+7) = 7/6

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doing the long division,

y = (2x^2-3x-17)/(2x^2-3x+7)
= 1 - 24/(2x^2-3x+7)

y' = 24(4x-3)/(2x^2-3x-7)^2

y'(-1) = 24(-7)/12^2 = -7/6
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using the quotient rule,

y = (2x^2-3x-17)/(2x^2-3x+7)
y' = [(4x-3)(2x^2-3x+7)-(2x^2-3x-17)(4x-3)]/(2x^2-3x+7)^2
= 24(4x-3)/(2x^2-3x+7)^2

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The tangent line is thus y+1 = -7/6 (x+1)

Check the graphs at

http://www.wolframalpha.com/input/?i=plot+(2x%5E2-3x-17)%2F(2x%5E2-3x%2B7),+y%3D(-7%2F6)(x%2B1)-1