Tangent line to a curve

Question

Find the Equation in point-slope form of the line tangent to the curve of x^2y-3y+2x at the point (2,5)

Answer 1

Assuming the equation is:

x²y - 3y + 2x = 9 [the point (2,5) does lie on that curve]

Use implicit differentiation to find and evaluate y'(2,5).

x²y' +2xy -3y' + 2 = 0

At this point we could solve for y', then substitute x=2, y = 5. OR you can just substitute those values now:

4y' +2(10) - 3y' + 2 = 0

y' + 22 = 0

y'(2,5) = -22

Using point-slope:

y - 5 = -22 (x - 2)

y = -22x + 44 + 5

y = -22x + 49 is the equation of the tangent line to the graph of the original function at the point (2,5).

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