Question

How do you write the quadratic in vertex form given `y=3x^2-12x+4`?

Answer 1

`y=3(x-2)^2-8`

Explanation:

vertex form is `y=a(x-h)^2 +k`

To solve this you have to complete the square with the x terms:

`y=3x^2-12x+4`

first isolate the x terms:

`y - 4=3x^2-12x`

`ax^2 +bx+c` to complete the square `a =1` and `c=(1/2b)^2`

so we need to factor out 3 so a = 1:

`y - 4=3(x^2-4x)`

now add the c to both sides, remember on the left side we need to multiply c by 3 since we are pulling out of the factor on the right:

`y - 4 +3c=3(x^2-4x +c)`

now solve for c:

`c=(1/2*-4)^2 = 4`

and put it into our equation and complete the square:

`y - 4 +3*4=3(x^2-4x +4)`

`y +8=3(x-2)^2`

finally isolate the y:

`y=3(x-2)^2-8`

given this form we know the vertex is `(-k, h) = (2, -8)` as shown in graph:

graph{y=3x^2-12x+4 [-8.05, 11.95, -8.84, 1.16]}