# What is the y-intercept of the line that is tangent to f(x) 4/(3x+1) at (-1, -2)? a) -5 b) -3/4 c) 3/4 d) -2 I got -3 as answer, however, it is not the answer in the choice.

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You need to find the equation of tangent line to the curve `f(x)=4/(3x+1), ` at the point (-1,-2), hence, you need to differentiate f(x) with respect to x using quotient rule such that:

`f'(x) = (4'*(3x+1) - 4*(3x+1)')/((3x+1)^2) `

`f'(x) = (-12)/((3x+1)^2)`

You need to find f(-1) and f'(-1) such that:

`f(-1) = 4/(-3+1) =gt f(-1) = -2`

`f'(-1) = (-12)/((-3+1)^2) =gt f'(-1) = -12/4 =gt f'(-1) = -3`

Hence, the equation of tangent line to the curve at x=-1 is:

`y - f(-1) = f'(-1)(x - (-1))`

Substituting -2 for f(-1) and -3 for f'(-1) yields:

`y + 2 = -3(x+1)`

Opening the brackets yields:

`y = -3x - 3 - 2 =gt y = -3x - 5`

You need to remember that the line `y = -3x - 5` intercepts y axis at x=0, hence y = -5.

**Hence, evaluating the y intercept of the tangent line to the curve `f(x)=4/(3x+1)` at the point (-1,-2) yields (0,-5).**