# For which point(s) on the graph of f(x)=5x/(x+2) is the slope of the tangent 2/5?I got (3,3) and (-7,7), is it correct?

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You need to remember that derivative of a function expresses the slope of the tangent line to graph of function, hence you need to differenatiate the function with respect to x and then you need to solve the equation `f'(x)=2/5` such that:

`f'(x) = ((5x)'*(x+2) - 5x*(x+2)')/((x+2)^2)`

`f'(x) = (5(x+2) - 5x)/((x+2)^2)`

Opening the brackets to numerator yields:

`f'(x) = (5x + 10 - 5x)/((x+2)^2) =gt f'(x) = 10/((x+2)^2)`

You need to solve the equation `f'(x)=2/5` such that:

`10/((x+2)^2) = 2/5`

You need to divide by 2 both sides such that:

`5/((x+2)^2) = 1/5 =gt (x+2)^2 = 25`

Expanding the binomial yields:

`x^2 + 4x + 4 - 25 = 0 =gt x^2 +4x - 21 = 0`

You need to use quadratic formula such that:

`x_(1,2) = (-4+-sqrt(16 + 84))/2 =gt x_(1,2) = (-4+-sqrt100)/2`

`x_(1,2) = (-4+-10)/2 =gt x_1 = 3 ; x_2 = -7`

You need to evaluate f(3) and f(-7) such that:

`f(3) = (5*3)/(3+2) =gt f(3) = 15/5 = 3`

`f(-7) = (-7*5)/(-7+2) = -35/(-5)=7`

**Hence, evaluating the points on the graph where slope of tangent is 2/5 yields (3;3) and (-7;7).**