Find the equation of the tangent line to the graph of f(x)= x-2 / x+1 at the point which is x=0.
-
You would use differentiation. The value derivative of a function f(x) at a certain point gives you the slope of the tangent at that point. So first we need to find the derivative:
y = (x-2)/(x+1) we derive as a quotient
y' = [(x-2)(x+1)' - (x-2)'(x+1)]/(x+1)^2 remember that y' = dy/dx meaning the derivative of y respect to x. Also remember, ax' = a, and constant' = 0
y' = (x-2 - (x+1))/(x^2+2x+1)
y' = -3/(x^2+2x+1) input x=0 to find the slope of the tangent of the function
y'(3) = -3
Now you know that the tangent has a slope of -3. To know the actual equation of the tangent you would need a point that belongs to the tangent, which you have not given. You use the equation:
y = m(x-a) + b where m is the slope (m=-3), a is the x coordinate of the point, and b is the y coordinate of the point. And that is your equation. Hope you understood!
y = (x-2)/(x+1) we derive as a quotient
y' = [(x-2)(x+1)' - (x-2)'(x+1)]/(x+1)^2 remember that y' = dy/dx meaning the derivative of y respect to x. Also remember, ax' = a, and constant' = 0
y' = (x-2 - (x+1))/(x^2+2x+1)
y' = -3/(x^2+2x+1) input x=0 to find the slope of the tangent of the function
y'(3) = -3
Now you know that the tangent has a slope of -3. To know the actual equation of the tangent you would need a point that belongs to the tangent, which you have not given. You use the equation:
y = m(x-a) + b where m is the slope (m=-3), a is the x coordinate of the point, and b is the y coordinate of the point. And that is your equation. Hope you understood!