Question

How do you find the vertex of `y = 10x – 3x^2`?

Answer 1

Using a 'sort of' cheat. Not really! It is part of the process of completing the square.

Vertex`->(x,y)=(5/3,25/3)`

Explanation:

Note that as `-3x^2` is negative the graph is of shape `nn`

Write as `y=-3x^2+10x+0`

This is now in the standardised form of

`y=ax^2+bx+c`

Change this to:

`y=a(x^2+b/ax)+c`

Note that `axxb/a=b`

`x_("vertex")=(-1/2)xxb/a`

`x_("vertex")=(-1/2)xx10/(-3) = +10/6 = 5/3`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute `x_("vertex")=+5/3` giving:

`y_("vertex")=-3(5/3)^2+10(5/3)`

`y_("vertex")=-75/9+50/3`

`y_("vertex")=-75/9+150/9`

`y_("vertex")=+75/9=8 1/3 ->25/3`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Vertex`->(x,y)=(5/3,25/3)`