# Can you find... cubic curves

Can you find equations for cubic curves that have specific features?

## Problem

Can you find a cubic curve that...

(a) ... has no stationary points?

(b) ... has two stationary points: one when $x=2$ and one when $x=5$?

(c) ... has a local minimum when $x=-1$?

(d) ... has a local minimum when $x=-2$ and a local maximum when $x=4$?

It would be a good idea to try and *sketch* some of the cubics first before trying to form an equation.

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## Student Solutions

Kieran Fitzgerald fro St Stephens Carramar in Australia got us started on the first parts of the problem using an algebraic approach:

a) For the cubic to have no stationary points the derivative cannot equal $0$

($dy/dx \neq 0$).This means that the discriminant must equal less than $0$. For

this to occur, assuming $y=ax^3+bx^2+cx+d$, both $a$ and $c$ must be either + or -

but they cannot be different. $b$ must also be considerably lower than $a$ and

$c$ as $b^2$ must be less than $4ac$. An example would be $y=4x^3-2x^2+3x$.

b) For a cubic to have stationary points at $x=2$ and $5$ the derivative must

equal $(x-2)(x-5)$. This can then be expanded and anti-differentiated to

become $y=x^3/3-3.5x^2+10x+d$ however $d$ can be any number.

c) In this case the derivative must equal $(x+1)(x+d)$. This is because there

will be a stationary point at $x= -1$ and another at $d$. To make $x=-1$ the

local minimum $d$ must be more than $1$ so that the other turning point is a

maximum. An example would be $y=x^3/3+2x^2+3x$.

Thomas from BHASVIC Sixth Form College Brighton used similar reasoning on the last part of the question:

d) We require $\frac{dy}{dx}=3(x+2)(x-4)$ for one max and one min at $x=-2,4$ in some order. Then integrating, $y=x^3-3x^2-24x$. This has the min and max the wrong way round so use transformation $y \rightarrow -y$ to reflect in $x$-axis, giving $$y=-x^3+3x^2+24x$$ as a possible cubic.

a) For the cubic to have no stationary points the derivative cannot equal $0$

($dy/dx \neq 0$).This means that the discriminant must equal less than $0$. For

this to occur, assuming $y=ax^3+bx^2+cx+d$, both $a$ and $c$ must be either + or -

but they cannot be different. $b$ must also be considerably lower than $a$ and

$c$ as $b^2$ must be less than $4ac$. An example would be $y=4x^3-2x^2+3x$.

b) For a cubic to have stationary points at $x=2$ and $5$ the derivative must

equal $(x-2)(x-5)$. This can then be expanded and anti-differentiated to

become $y=x^3/3-3.5x^2+10x+d$ however $d$ can be any number.

c) In this case the derivative must equal $(x+1)(x+d)$. This is because there

will be a stationary point at $x= -1$ and another at $d$. To make $x=-1$ the

local minimum $d$ must be more than $1$ so that the other turning point is a

maximum. An example would be $y=x^3/3+2x^2+3x$.

Thomas from BHASVIC Sixth Form College Brighton used similar reasoning on the last part of the question:

d) We require $\frac{dy}{dx}=3(x+2)(x-4)$ for one max and one min at $x=-2,4$ in some order. Then integrating, $y=x^3-3x^2-24x$. This has the min and max the wrong way round so use transformation $y \rightarrow -y$ to reflect in $x$-axis, giving $$y=-x^3+3x^2+24x$$ as a possible cubic.