Math  /  Calculus

QuestionFind equations of the tangent lines to the curve y2xy12=0y^{2}-x y-12=0 at the points (1,3)(-1,3), and (1,4)(-1,-4).
The tangent line at (1,3)(-1,3) is y=y= \square (Type your answer in slope-intercept form. Use integers or fractions for any numbers in the expression.)

Studdy Solution
(1,3)(-1, 3) noktasındaki teğet doğrusu: y=37x+247y = \frac{3}{7}x + \frac{24}{7}
(1,4)(-1, -4) noktasındaki teğet doğrusu: y=47x247y = \frac{4}{7}x - \frac{24}{7}

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