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long live @geerky42
Normal is perpendicular to tangent line so normal at f(3) is perpendicular to tangent line at f(3). Given that normal is \(\dfrac{3\pi}{4}\) (aka 135 deg) away from x-axia, you just need to find the slope of line perpendicular to normal (90 deg away from 135 deg (aka 45 deg))
|dw:1397168518404:dw|
Make sense? @helpasap
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Let say slope for line N is m. For line T, which is perpendicular to line N, the slope is -1/m. Hope you can figure it out.
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