please make sure it's easy to read Find the points at which the graph of the equation has a vertical or horizontal tangent line.16x² + y²32x + 8y + 16 = 0-horizontal tangentsvertical tangents(x, y) = (0,(x, y) =(x, y) =(x, y) =- 42, - 41,01,8XX(smaller y-value)(larger y-value)(smaller x-value)(larger x-value)

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Find the points at which the graph of the equation has a vertical or horizontal tangent line. 16x2 + y² 32x + 8y + 16 = 0 horizontal tangents (x, y) = (0, -4 (x, y) = (2,- (x, y) = vertical tangents - (x, y) = 1,0 4 1,8 X X X X ) (smaller y-value) ) (larger y-value) ) (smaller x-value) (larger x-value)