let f(x)= 2/1+x^2. The equation of the tangent line to the curve at the point (5,0.07692) can be written in the form y=mx+b where m is: ? and b is?

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let f(x)= 2/1+x^2. The equation of the tangent line to the curve at the point (5,0.07692) can be written in the form y=mx+b where m is: ? and b is?

Mathematics
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m = 2x at x=5
y-y1=m(x-x1)
y1 x1 are your points plug, rearrange, and determine your values

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point-slope form for algebra is this:\[y-y_1=m(x-x_1)\]now in calculus we learn that slope of the tangent at a point 'a' is the derivative at that point\[m=f'(a)\]so we need the coordinates of the point\[(x_1,y_1)\]and the value of the derivative \[f'(x_1)\]then we can use the modified point-slope form\[y-y_1=f'(x_1)(x-x_1)\]
rearrange into slope intercept form...

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