Please check if this is a true statement. In order to aproximate a tangent line, I would take the dervitive and pick a known point (x,y) then sub in my x value. then put in into point slope right? Asking because i have to write 3 questions and provide answers
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by "approximate" a tangent line are you wnating to find the equation that will match the tangent you get from the derivative?
if you have a function f(x) and want to know the equation of the tangent line at any given point:
then y = f'(x)(x) + b will suffice.
That is what i am guessing he ment by it but i am not 100% sure he seems to be kinda cryptic at times. I was kinda wondering if it's a standard operation and if so thats how you would figure it out.
suppose we take f(x) = 3x^2
f'(x) = 6x at any point given and will produce the slope of the tangent.
this f'(x) substitutes into the normal equation of the line and you use the (x,y) point as the values to calibrate your new equation.
sounds good thanks for the help
so the equation of the tangent at (4,48) becomes:
48 = 6(4)(4) + B
48 = 96 + B
-48 = B
y = (6x)x - 48
for that particular point at (4,48).
youre welcome :)