1. ## foci on parabola

The parabola $y = x^2+3x$ has its focus at the point (b, c) where
b=
c=
i tried using the stardard equation $y^2 = 4px$, but got lost:
alright i did this by completing the square:
$x^2+3x = y$
$x^2+3x+\frac{9}{4} = y + \frac{9}{4}$
$(x+\frac{3}{2})^2 = y+\frac{9}{4}$
so $b = \frac{-3}{2}$, need help finding c

2. The focus is a distance p from the vertex. $p=\frac{1}{4a}$

Where a=1. So p=1/4. Since the vertex is at (-3/2, -9/4). The parabola opens upward, so the focus will be above the vertex. -9/4+1/4=-2

F(-3/2, -2)