Question

Where does the function x^3+y^2−12x+16y=28 have a horizontal tangent line

Answer 1

The function has a horizontal tangent line at (2.45, -18.33), (2.45, 2.33) , (-2.45, -16.79) and (-2.45, 0.79)

The function x³ + y² − 12x + 16y = 28 has a horizontal tangent when its derivative dy/dx = 0.

So, differentiating implictly, we have

d(x³ + y² − 12x + 16y)/dx = d28/dx

dx³/dx + dy²/dx − d12x/dx + d16y/dx = d28/dx

2x² + 2ydy/dx - 12 + dy/dx = 0

(2y + 1)dy/dx = 12 - 2x²

dy/dx = (12 - 2x²)/(2y + 1)

So, the function has a horizontal tangent line when dy/dx = 0

So,  (12 - 2x²)/(2y + 1) = 0

12 - 2x² = 0

2x² = 12

x² = 12/2

x² = 6

taking square-root of both sides, we have

x = ±√6

Substituting x = √6 = 2.45 into the equation, we have

x³ + y² − 12x + 16y = 28

(√6 )³ + y² − 12(√6) + 16y = 28

6√6  + y² − 12√6 + 16y = 28

y² + 16y + 6√6 − 12√6 - 28 = 0

y² + 16y - (6√6 + 28) = 0

Using the quadratic formula, we find y

So,

$y = \frac{-16 +/- \sqrt{16^{2} - 4 X 1 X -(6\sqrt{6} + 28)} }{2 X 1} \\y = \frac{-16 +/- \sqrt{256+ 24\sqrt{6} + 112)} }{2} \\y = \frac{-16 +/- \sqrt{368 + 24\sqrt{6})} }{2}\\y = \frac{-16 +/- \sqrt{368 + 24X2.45} }{2}\\y = \frac{-16 +/- \sqrt{368 + 58.79} }{2}\\y = \frac{-16 +/- \sqrt{426.79} }{2}\\y = \frac{-16 +/- 20.66 }{2}\\y = \frac{-16 - 20.66 }{2} or y = \frac{-16 + 20.66 }{2}\\y = \frac{-36.65 }{2} or y = \frac{4.66 }{2}\\y = -18.33 or 2.33$

Substituting x = -√6 = -2.45 into the equation, we have

x³ + y² − 12x + 16y = 28

(-√6 )³ + y² − 12(-√6) + 16y = 28

-6√6  + y² + 12√6 + 16y = 28

y² + 16y - 6√6 + 12√6 - 28 = 0

y² + 16y + (6√6 - 28) = 0

Using the quadratic formula, we find y

So,

$y = \frac{-16 +/- \sqrt{16^{2} - 4 X 1 X (6\sqrt{6} - 28)} }{2 X 1} \\y = \frac{-16 +/- \sqrt{256- 24\sqrt{6} + 112)} }{2} \\y = \frac{-16 +/- \sqrt{368 - 24\sqrt{6})} }{2}\\y = \frac{-16 +/- \sqrt{368 - 24X2.45} }{2}\\y = \frac{-16 +/- \sqrt{368 - 58.79} }{2}\\y = \frac{-16 +/- \sqrt{309.21} }{2}\\y = \frac{-16 +/- 17.58 }{2}\\y = \frac{-16 - 17.58 }{2} or y = \frac{-16 + 17.58 }{2}\\y = \frac{-33.58 }{2} or y = \frac{1.58 }{2}\\y = -16.79 or 0.79$

So, the function has a horizontal tangent line at (2.45, -18.33), (2.45, 2.33) , (-2.45, -16.79) and (-2.45, 0.79)

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