Answer
a. $y=x^3+1$
b. Only one curve.
Work Step by Step
a. Given $y''=6x$, we have $y'=3x^2+C$. As $y'(0)=0$ (horizontal tangent), we have $3(0)^2+C=0$ and $C=0$. Thus $y'=3x^2$ and we have $y=x^3+D$. Use the point $(0,1)$ on the curve to get $(0)^3+D=1$ and $D=1$ which leads to the answer $y=x^3+1$
b. Because we have found all the parameters $C$ and $D$, the function is unique, which means that there is only one curve.
