Question

What is the vertex form of `y=(x+4)(2x-1)(x-1)`?

Answer 1

Something like:

`f(x) = 2(x+5/6)x^3 - 91/6(x+5/6)+418/27`

Explanation:

The given polynomial is a cubic, not a quadratic. So we cannot reduce it to 'vertex form'.

What is interesting to do is to find a similar concept for cubics.

For quadratics we complete the square, thereby finding the centre of symmetry of the parabola.

For cubics we can make a linear substitution "completing the cube" to find the centre of the cubic curve.

`108 f(x) = 108(x+4)(2x-1)(x-1)`

`color(white)(108f(x)) = 108(2x^3+5x^2-11x+4)`

`color(white)(108f(x)) = 216x^3+540x^2-1188x+432`

`color(white)(108f(x)) = (6x)^3+3(6x)^2(5)+3(6x)(5)^2+(5)^3 -273(6x)-273(5)+1672`

`color(white)(108f(x)) = (6x+5)^3-273(6x+5)+1672`

So:

`f(x) = 1/108 (6x+5)^3 - 91/36(6x+5)+418/27`

`color(white)(f(x)) = 2(x+5/6)^3 - 91/6(x+5/6)+418/27`

From this we can read that the centre of symmetry of the cubic is at `(-5/6, 418/27)` and the multiplier `2` tells us that it is essentially twice as steep as `x^3` (though the linear term subtracts a constant `91/6` from the slope).

graph{(y-(x+4)(2x-1)(x-1))(40(x+5/6)^2+(y-418/27)^2-0.2) = 0 [-6.13, 3.87, -5, 40]}

So in general we can use this method to get a cubic function into the form:

`y = a(x-h)^3+m(x-h)+k`

where `a` is a multiplier indicating the steepness of the cubic compared with `x^3`, `m` is the slope at the centre point and `(h, k)` is the centre point.