I am sure you meant
f(t) = t^2 - 1
f '(t) = 2t
when t = 5
f(5) = 24
so we have the contact point (5,24)
f '(5) = 10
so the slope of the line is 10
finish it by using the method you learned to find the equation of a line, given the slope and a point
f(t) = t2 − 1
Find the equation of the line tangent to the graph of f(t) at t = 5.
Enter the equation of the tangent line here (in terms of the variable t):
y =
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