Challenge level

Use this proof sorter to construct a proof that it is impossible to find a cubic equation which passes through $(0,0)$ and has turning points at $(1,2)$ and $(2,1)$. This particular type of proof is called a *proof by contradiction*.

There are several different correct arrangements of these statements. Once you have had a go at rearranging the statements you can compare your answer to the possible solution which can be seen at the bottom of the page by clicking on the "Show Possible Solution" button.

See Curve Fitter for the problem that this proof sorter relates to.

Start by assuming that there is a cubic with turning points at $(1,2)$ and $(2,1)$, and which goes through $(0,0)$.

A general cubic has the form $y=ax^3+bx^2 + cx + d$.

The curve passes through $(0,0)$ and so we have $d=0$.

The curve passes through $(1,2)$ and substituting these values gives $2=a+b+c \quad (*)$.

The curve passes through $(2,1)$ and substituting these values gives $1=8a+4b+2c \quad (**)$.

The derivative of the cubic is $\frac {\text{d} y}{\text{d} x} =3ax^2+2bx+c$.

There is a turning point when $x=1$ so using the derivative equation we have $0=3a+2b+c \quad (***)$.

Eliminating $c$ by using $(**)-2(*)$ gives $-3=6a+2b \quad (\dagger)$.

Eliminating $c$ by using $(***)-(*)$ gives $-2=2a+b \quad (\ddagger)$.

Eliminating $b$ by using $(\dagger)-2(\ddagger)$ gives $1=2a \implies a=\frac 1 2$.

Substituting $a=\frac 1 2$ into $(\ddagger)$ gives $b=-3$.

Substituting $a=\frac 1 2$ and $b=-3$ into $(*)$ gives $c=\frac 9 2$.

The equation of the cubic is now $y=\frac 1 2 x^3 - 3x^2 + \frac 9 2 x$.

The derivative of the cubic is $\frac {\text{d} y}{\text{d} x} =\frac 3 2 x^2 -6x+\frac 9 2$.

When $x=2$ the derivative is $\frac 3 2 \times 2^2 - 6 \times 2 + \frac 9 2 = -\frac 3 2 $.

Hence the point $(2, 1)$ is

**not**a turning point of the cubic.

Therefore you cannot find a cubic which passes through $(0,0)$ and has turning points at $(1,2)$ and $(2,1)$.