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nzrodrigue nzrodrigue
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11 years ago
I need to calculate an equation for the tangent in the point S where x = 4 for the function (2x+1)/(x-3).
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wrote...
#1
11 years ago
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
wrote...
#2
11 years ago
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.
wrote...
#3
11 years ago
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
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