Integral calculus

A cubic function has a y intercept at (0,2) when x = 1, y = 3 , dy/dx = 2 and d^2y/dx^2 = 0. Find the equation of the original curve and sketch the curve

You don't know the cubic function so let's say

$y = ax^{3} + bx^{2} + cx + d$ (1).

Now, before we differentiate, let's try and find what we know.

Firstly, the y intercept is (0,2) so substituting this into (1) gives you $2 = 0 + 0 + 0 + d \Rightarrow d = 2$.

We also know that (1,3) is a solution of the cubic equation, therefore substituting this into (1) and with $d=2$,

$3 = a + b + c + 2 \Rightarrow 1 = a + b + c$ (2)

Now, the differentiating part. We're given the value of $y'$ and $y''$

$\displaystyle \frac{dy}{dx} = 3ax^{2} + 2bx + c$ (3)

$\displaystyle \frac{d^{2}y}{dx^{2}} = 6ax + 2b$ (4)

We know that $y' = 2$ and $y'' = 0$ both at $x=1$ so equating and subbing them both into (3) and (4),

$3a + 2b + c = 2$ (5)

$6a + 2b = 0 \Rightarrow b=-3a$ (6)

(To get (6), I divided the whole equation by 2 then rearranged to simplify it)

So here's what you have so far

$3a + 2b + c = 2$ (5)

$a + b + c = 1$ (2)

$b=-3a$ (6)

Now, simultaneous substitution (or elimination but it's a bit hard typing it up).

(6) into (5), $3a + 2(-3a) + c = 2 \Rightarrow c = 2 + 3a$ (7)

Now, substitute (7) and (6) into (2)

$a + (-3a) + (2+3a) = 1 \Rightarrow a = -1$

Sub $a=-1$ into (6) and (7) to get $b=3$ and $c=-1$ respectively.

So subbing everything back in (1), you get

$y=-x^{3}+3x^{2}-x+2$.

Now you should always check your values of a, b and c and likewise check the cubic equation if it matches the values given from the question, so you differentiate my answer and sub in $x=1$ into y' and y'' and you'll get the correct values y' = 2 and y'' = 0 .

To sketch the curve, find the intercept points in the x and y axis along with the stationary points, use the double differentation method to find if the point is max or min.