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

Feb 18, 2018

Something like:

$f \left(x\right) = 2 \left(x + \frac{5}{6}\right) {x}^{3} - \frac{91}{6} \left(x + \frac{5}{6}\right) + \frac{418}{27}$

#### 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.

$108 f \left(x\right) = 108 \left(x + 4\right) \left(2 x - 1\right) \left(x - 1\right)$

$\textcolor{w h i t e}{108 f \left(x\right)} = 108 \left(2 {x}^{3} + 5 {x}^{2} - 11 x + 4\right)$

$\textcolor{w h i t e}{108 f \left(x\right)} = 216 {x}^{3} + 540 {x}^{2} - 1188 x + 432$

$\textcolor{w h i t e}{108 f \left(x\right)} = {\left(6 x\right)}^{3} + 3 {\left(6 x\right)}^{2} \left(5\right) + 3 \left(6 x\right) {\left(5\right)}^{2} + {\left(5\right)}^{3} - 273 \left(6 x\right) - 273 \left(5\right) + 1672$

$\textcolor{w h i t e}{108 f \left(x\right)} = {\left(6 x + 5\right)}^{3} - 273 \left(6 x + 5\right) + 1672$

So:

$f \left(x\right) = \frac{1}{108} {\left(6 x + 5\right)}^{3} - \frac{91}{36} \left(6 x + 5\right) + \frac{418}{27}$

$\textcolor{w h i t e}{f \left(x\right)} = 2 {\left(x + \frac{5}{6}\right)}^{3} - \frac{91}{6} \left(x + \frac{5}{6}\right) + \frac{418}{27}$

From this we can read that the centre of symmetry of the cubic is at $\left(- \frac{5}{6} , \frac{418}{27}\right)$ and the multiplier $2$ tells us that it is essentially twice as steep as ${x}^{3}$ (though the linear term subtracts a constant $\frac{91}{6}$ 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:

$y = a {\left(x - h\right)}^{3} + m \left(x - h\right) + k$

where $a$ is a multiplier indicating the steepness of the cubic compared with ${x}^{3}$, $m$ is the slope at the centre point and $\left(h , k\right)$ is the centre point.