Slope of the tangent line

• Jul 19th 2009, 02:47 PM
xterminal01
Slope of the tangent line
The slope of the tangent live to a curve is given by f(x)=3x^2+5 if the point
(1,2) is on the curve, find the equation of the curve.
if i integrate this i get x^3+5x+C however the result has a constant of -4, that is => x^3+5x-4
So how would i go about finding the constant ?
• Jul 19th 2009, 02:56 PM
skeeter
Quote:

Originally Posted by xterminal01
The slope of the tangent live to a curve is given by f(x)=3x^2+5 if the point
(1,2) is on the curve, find the equation of the curve.
if i integrate this i get x^3+5x+C however the result has a constant of -4, that is => x^3+5x-4
So how would i go about finding the constant ?

plug in the initial condition ...

$f(x) = x^3 + 5x + C$

$f(1) = 1^3 + 5(1) + C$

$2 = 6 + C$

$C = -4$
• Jul 19th 2009, 02:57 PM
VonNemo19
Quote:

Originally Posted by xterminal01
The slope of the tangent live to a curve is given by f`(x)=3x^2+5 if the point
(1,2) is on the curve, find the equation of the curve.
if i integrate this i get x^3+5x+C however the result has a constant of -4, that is => x^3+5x-4
So how would i go about finding the constant ?

(1,2) <---this is your initial condition

$(1,2)\text{ on f(x) }\Rightarrow2=(1)^3+5(1)+C$