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PLEASE HELP!!! MEDAL WILL BE REWARDED! Find the points at which the graph of the equation has a vertical or horizontal tangent line. 4x^2+y^2-8x+4y+4=0
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4x² + y² - 8x + 4y + 4 = 0 or 8x + 2y dy/dx - 8 + 4 dy/dx = 0 or (8x - 8) + (2y + 4) dy/dx = 0 or dy/dx = - 4 (x - 1) / (y + 2) Horizontal Tangent Line: dy/dx = 0 ... → x = 1 ← final solution Vertical Tangent Line: dx/dy = 0 ...... → y = -2 ← final solution
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