Equation of line normal to curve

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.

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

Quote:

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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.

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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?