## Algebra 1: Common Core (15th Edition)

vertex: $( -6,-504)$
Consider the equation: $y=x^2+12x-468$ ...(1) The standard equation of the parabola is: $y=a(x-h)^2+k$ with vertex$(h,k)$ Consider right side $x^2+12x=468$ To complete the square add $(\frac{b}{2})^2=(\frac{12}{2})^2=36$. $x^2+12x+36=468+36$ $(x+6)^2=504$ Thus, equation (1) becomes: $y=(x+6)^2-504$ Compare it with the standard form of the parabola, we get vertex: $( -6,-504)$