Find an equation of the curve $y=f(x)$

Question

$y'' = 6x$ and the tangent line to the curve at $(1,2)$ is horizontal

Can anybody help me solve this? How am I supposed to approche this question?

Answer 1

As the comment from @David Peterson states, you are looking for a function that verifies $y(1)=2$ (i.e. the curve $y=y(x)$ passes through $(1,2)$) and $y'(1)=0$ (the curve is horizontal at this point)), and also verifies your differential equation.

I assume you know how to integrate and that you are integrating w.r.t. the variable $x$, considering $y$ as a function of $x$. This gives $$y(x)= x^3+ax+b$$ for some constant values $a,b$. In order to satisfy your constraints, they must verify $$2=1+a+b\mbox{ and }0=3+a$$ i.e. $a=-3,b=4$. Your function is $y(x)=x^3-3x+4$.

Answer 2

If $$y''=6x$$ then $$y'=3x^2 + C_1$$ and $$y=x^3 + C_1 x + C_2$$

Tangent line is horizontal at $(1,2)$ means that the derivative of your curve is 0 at 1.

$$0=3.1^2 + C_1 \iff C_1=-3$$

The final information to use is that $y(1)=2$

$$2= 1^3 -3 + C_2 \iff C_2=4$$

So your curve is

$$y=x^3 - 3x + 4$$