Asked by Angelica
Find the equation of the tangent line to the graph of
f(x) =
2x − 6/x + 1
at the point at which
x = 0.
(Let x be the independent variable and y be the dependent variable.)
I don't know how to find the derivative, i Keep getting it wrong. But I know where to go after that.
f(x) =
2x − 6/x + 1
at the point at which
x = 0.
(Let x be the independent variable and y be the dependent variable.)
I don't know how to find the derivative, i Keep getting it wrong. But I know where to go after that.
Answers
Answered by
Steve
Assuming you meant
f = (2x-6)/(x+1)
using the quotient rule,
f' = [(2)(x+1)-(2x-6)(1)]/(x+1)^2
so, at x=0, f' = (2+6)/2 = 4
Or, you can note that
f = (2x-6)/(x+1)
= 2 - 8/(x+1)
so,
f' = 8/(x+1)^2
f'(0) = 8/2 = 4
f = (2x-6)/(x+1)
using the quotient rule,
f' = [(2)(x+1)-(2x-6)(1)]/(x+1)^2
so, at x=0, f' = (2+6)/2 = 4
Or, you can note that
f = (2x-6)/(x+1)
= 2 - 8/(x+1)
so,
f' = 8/(x+1)^2
f'(0) = 8/2 = 4
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