In order to find the equation of a tangent line you must take the derivative of the original equation and then find the points that it passes through.
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The question is suppose to read: Find the equation of the line tangent to y=(x²+3x)²(2x-2)³, when x=8
Slope is the tangent of the angle between a given straight line and the x-axis of a system of Cartesian coordinates
Tangent to orbit, north away, tangent to orbit, south away.
Points: (7, 3) (14, 6) (21, 9) Slope: 3/7 Equation: 7y=3x
x + 2y = 7 so x = 7 - 2ySubstitute in the equation of the circle:(7 - 2y)^2 + y^2 - 4*(7 - 2y) - 1 = 049 - 28y + 4y^2 + y^2 - 28 + 8y - 1 = 0=> 5y^2 - 20y + 20 = 0=> y^2 - 4y + 4 = 0 => y = 2and therefore x = 7 - 2y = 3Thus the straight line intersects the circle at only one point (3, 2) and therefore it must be a tangent.