Question

Find the equation of a curve going through point (3,7) and having slope 4x^2-3 at (x,y)

Answer 1

So far this is what I have.


\[\frac{ dy }{ dx } = 4x^{2}-3\]

\[dy = (4x^{2}-3)~dx\]

\[\int\limits dy = \int\limits (4x^{2}-3)~dx\]

\[y = \frac{ 4 }{ 3 }x^{3}-3x+C\]

Answer 2

y=(4/3)x^3-3x+c
Substitute (x,y)=(3,7) and solve for C to find the constant.

Answer 3

(3,7) to find C

Answer 4

oh wow haha .. that's it I over thought this

Answer 5

Okay, I see now

\[\frac{ 4 }{ 3 }(3)^{3}-3(3)+C = 7\]

Answer 6

27+C = 20

So then this would make C = -7

Answer 7

don't you mean
27+C = 7 ?

Answer 8

got disconnected

Answer 9

Thanks for clarifying

27+C = 7

So this would make C = -20

=> C = -20

Answer 10

yes

Answer 11

Good job @Photon336 !

Answer 12

thank you @mathmate

Answer 13

You're welcome!