when x=10 , f(10)=500
so your point is (10,500)
f'(x) = 10x
then when x=10 , slop = 100
Now you have a slope of 100 and a point of (10,500)
Use the method you learned to find the equation of the tangent (a straight line)
find an equation of the line tangent to the equation at the given point.
f(x)=5x^2 at x=10
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