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1 answer:
The point-slope form of the equation for a line can be written as
... y = m(x -h) +k . . . . . . . for a line with slope m through point (h, k)
Your function gives
... f'(h) = m
... f(h) = k
a) The tangent line is then
... y = 5(x -2) +3
b) The normal line will have a slope that is the negative reciprocal of that of the tangent line.
... y = (-1/5)(x -2) +3
_____
You asked for "an equation." That's what is provided above. Each can be rearranged to whatever form you like.
In standard form, the tangent line's equation is 5x -y = 7 . The normal line's equation is x +5y = 17 .
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In order to be perpendicular, you must have a slope that is the opposite reciprocal, so the slope should by -3/4. So the equation could be y = -3/4x with any other number at the end.