- #1
steven10137
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Homework Statement
The tangent to the curve [tex]y=f(x)[/tex] at the point (0,1) has equation [tex]y=x+1[/tex]. The gradient function of the curve is given by [tex]\frac{{dy}}{{dx}} = ax + b [/tex]. Find the equation of the curve given that it also passes through the point (1,3)
2. The attempt at a solution
Well we have the curve y=f(x) which I am assuming to be a quadratic or something of the like, and at the point (0,1) it has the tangent equation [tex]y=x+1[/tex].
The gradient function dy/dx, at the point (0,1):
[tex]\frac{{dy}}{{dx}} = ax + b [/tex]
[tex]\therefore[/tex] [tex]1 = a(0) + b[/tex]
hence b=1
Can we now solve for 'a' by saying that:
[tex]ax+1 = x+1[/tex]
??
from that a=1 and [tex]y=x+1[/tex] as before.
But the integral of [tex]y=x+1[/tex] is [tex]y=1/2 x^2 +x + C[/tex]
@ (1,3)
c=1/2
[tex]y=1/2 x^2 +x + 1/2[/tex]
The answer is:
[tex]y=1/2 x^2 +x + 1[/tex]
Where is my mistake?!?
thanks
Steven