## Intermediate Algebra for College Students (7th Edition)

The vertex of $x=a(y-k)^2+h$ is at $(h,k)$. The vertex of $y=a(x-k)^2+h$ is at $(k,h)$. I modify the original equation: $y=x^2+6x+10\\y=x^2+6x+9+1\\y=(x+3)^2+1$ a) $x$ is squared, thus the parabola is vertical b) the coefficient of $x^2$ is positive, thus the parabola opens upwards c) the vertex is at $(-3,1)$.