## Calculus: Early Transcendentals 8th Edition

$x = \frac{1}{2}(y+3)^2-2$ This is the equation of a parabola with vertex $(-2,-3)$
$y^2-2x+6y+5 = 0$ $2x = y^2+6y+5$ $2x = y^2+6y+9-9+5$ $2x = (y+3)^2-4$ $x = \frac{1}{2}(y+3)^2-2$ This is the equation of a parabola with vertex $(-2,-3)$