Question

What is the vertex form of `y= 9x^2 + 27x + 27`?

Answer 1

The solution set is: `S={-3/2, -27/4}`

Explanation:

The general formula for a quadratic function is:
`y=Ax^2+Bx+C`

To find the vertex, we apply those formulas:
`x_(vertex)=−b/(2a)`
`y_(vertex)=−△/(4a)`

In this case:
`x_(vertex)= -(27/18) = -3/2`

`y_(vertex)= - (27^2 - 4 * 9 * 27)/(4*9)` To make it easier, we factor the multiples of 3, like this:
`y_(vertex)= - ((3^3)^2 - 4 * 3^2 * 3^3)/(4 * 3^2)`
`y_(vertex)= - (3^6 - 4 * 3^5) / (4 * 3^2) = (3^4 * cancel(3^2) -4*3^3*cancel(3^2))/(4 * cancel(3^2))`
`y_(vertex)= - (81 - 108)/4 = -27/4`

So, the solution set is: `S={-3/2, -27/4}`