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How would I write an equation for this tangent line, g(3)=6 , g'(3)=-2
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g(3)=6 means that when x = 3, y = 6 So the point (3,6) lies on the curve g(3) = -2 means that on the curve at x = 3, the tangent line has a slope of m = -2 So use the point slope formula y-y1 = m(x-x1) and plug these values in to get y-6 = -2(x-3) From here, solve for y to find the equation for the tangent line
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