# How to graph cubics?

## () For question a, can somebody please tell me how to get to the answer in the photo?

Apr 9, 2017

$y = {x}^{3} - 3 {x}^{2} + 3 x - 1$

#### Explanation:

The general equation of a cubic polynomial is: $y = a {x}^{3} + b {x}^{2} + c x + d$

From looking at the graph in the question labelled '(a)', it appears that:

$y = - 1$ at $x = 0$ (1)

$y = 0$ at $x = 1$ (2)

Slope of $y = 0$ at $x = 1$ (3)

$y$ has an inflection point at $x = 1$ (4)

Assuming (1) through (4) above are true:

$\left(1\right) \to d = - 1$ {A]

$\left(2\right) \to a + b + c + d = 0$ [B]

$y ' = 3 a {x}^{2} + 2 b x + c$

Hence $\left(3\right) \to 3 a + 2 b + c = 0$ [C]

$y ' ' = 6 a x + 2 b$

Hence $\left(4\right) \to 6 a + 2 b = 0$

$\therefore b = - 3 a$ [D}

[D] in [C} $\to 3 a - 6 a + c = 0$

$\therefore c = 3 a$ [E]

[A], [D] and [E] in [B] $\to a - 3 a + 3 a - 1 = 0$

Hence: $a = 1$ $\to b = - 3$ and $c = 3$

$\therefore$ our cubic is: $y = {x}^{3} - 3 {x}^{2} + 3 x - 1$