|
Title: How do I calculate the equation of a tangent in a certain point? Post by: nzrodrigue on Sep 27, 2012 I need to calculate an equation for the tangent in the point S where x = 4 for the function (2x+1)/(x-3).
Title: How do I calculate the equation of a tangent in a certain point? Post by: tommyboiazn23 on Sep 27, 2012 first take the derivative.
use quotient rule. (f 'g-g'f)/g^2 f=2x+1 g=x-3 f '=2 g'=1 (2(x-3)-(2x+1))/(x-3)^2 (2x-6-2x-1)/(x-3)^2 -7/(x-3)^2 then plug in the given value to find the slope of the tangent line -7/(4-3)^2 -7 then plug in a point on the original equation to point slope form and you have a tangent line (0,-1/3) is on the line by (2(0)+1)/(0-3) y+1/3=-7x make it a good day Title: How do I calculate the equation of a tangent in a certain point? Post by: nypr1nc355 on Sep 27, 2012 First, determine "S" by plugging 4 into the function. You get 9, so you're looking for the tangent line at (4,9).
Now, take a derivative the function. You get (2(x-3) - (2x+1)) / (x-3)^2. Substitute 4 into this equation to find the slope at (4,9). y = (2(x-3) - (2x+1)) / (x-3)^2 y = (2(1) - (9)) / (1)^2 y = -7 Finally, put the equation in point-slope format and sub in (x,y) to find the tangent line. y - y1 = m(x - x1) y - 9 = -7(x - 4) You can simplify it to y = -7x - 19 if you want. Title: How do I calculate the equation of a tangent in a certain point? Post by: lester on Sep 27, 2012 The eqn of tangent of the curve f(x) at x = a is given by
y = f (a) + f ' (a)(x - a) here f (x) = (2x + 1) / (x - 3) a = 4 f(4) = (2*4 + 1) / (4 - 3) = 9 f ' (x) = [ (x - 3)(2) - (2x + 1)(1) ] / (x - 3)^2 = [ 2x - 6 - 2x - 1 ] / (x - 3)^2 f ' (x) = - 7 / (x - 3)^2 f ' (4) = - 7 / 1 = -7 eqn of tangent is y = f(a) + f ' (a)(x - a) y = 9 - 7 (x - 4) y = 9 - 7x + 28 y = -7x + 37 |