Hi. Equation for curve: y = 6x - 2x^2 at (2,4) My work: = 6 - 4x = 6 - 4(2) = 6 - 8 = -2 = M (slope) reciprocal: 1/-2 y - y1 = m (x - x1) y - 4 = -2(x-2) y - 4 = -2x + 4 y = -2x but this is wrong? thanks.
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The normal line is the line perpindicular to the curve, and that requires the negative reciprocal I think. So the slope of the normal line should be just 1/2. You can type the question into WolframAlpha to confirm.
got it. thanks
Originally Posted by togo Hi. Equation for curve: y = 6x - 2x^2 at (2,4) My work: = 6 - 4x = 6 - 4(2) = 6 - 8 = -2 = M (slope) reciprocal: 1/-2 y - y1 = m (x - x1) y - 4 = -2(x-2) y - 4 = -2x + 4 y = -2x but this is wrong? thanks. It's already been established that the gradient of your normal is 1/2, not -1/2. But I'm wondering why you substituted -2 as m?
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