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Find the equation of the tangent line to the curve y
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Find the equation of the tangent line to the curve y =4 secx-8cosx at the point(pi/3,4). The equation of this tangent line can be written in the form y=mx+b
okay so I have \[m = 4(2)(\sqrt{3})+8(\frac{ \sqrt{3} }{ 2 })\]
how do I find what 'b' is?
a tangent line is usually of the point slopr format, not the slope intercept format
given a point (a,f(a)) and a slope f'(a) a point slope format is of the form: y-f(a) = f'(a) (x-a)
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this can be converted to slope intercept simply by expanding, and solving for y\ y = f'(a) x - a f'(a) + f(a)
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