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

May 20, 2018

$y = 3 {\left(x - 2\right)}^{2} - 8$

#### Explanation:

vertex form is $y = a {\left(x - h\right)}^{2} + k$

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

$y = 3 {x}^{2} - 12 x + 4$

first isolate the x terms:

$y - 4 = 3 {x}^{2} - 12 x$

$a {x}^{2} + b x + c$ to complete the square $a = 1$ and $c = {\left(\frac{1}{2} b\right)}^{2}$

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

$y - 4 = 3 \left({x}^{2} - 4 x\right)$

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:

$y - 4 + 3 c = 3 \left({x}^{2} - 4 x + c\right)$

now solve for c:

$c = {\left(\frac{1}{2} \cdot - 4\right)}^{2} = 4$

and put it into our equation and complete the square:

$y - 4 + 3 \cdot 4 = 3 \left({x}^{2} - 4 x + 4\right)$

$y + 8 = 3 {\left(x - 2\right)}^{2}$

finally isolate the y:

$y = 3 {\left(x - 2\right)}^{2} - 8$

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

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