## Precalculus (10th Edition)

Vertex: $(37,-6)$ Focus: $\left(36.75,-6\right)$ Directrix: $x=37.25$ See graph
We are given the parabola: $y^2+12y=-x+1$ Put the equation in standard form: $y^2+12y+36=-x+1+36$ $(y+6)^2=-(x-37)$ The standard equation is: $(y-k)^2=4p(x-h)$ Determine $h,k,p$: $h=37$ $k=-6$ $4p=-1\Rightarrow p=-0.25$ Determine the vertex: $(h,k)=(37,-6)$ Determine the focus: $(h+p,k)=\left(37+(-0.25),-6\right)=\left(36.75,-6\right)$ Determine the directrix: $x=h-p$ $x=37-(-0.25)$ $x=37.25$ Determine the two points defining the latus rectum: $x=36.75$ $(y+6)^2=-36.75+37$ $(y+6)^2=0.25$ $y+6=\pm 0.5$ $y+6=-0.5\Rightarrow y_1=-6.5$ $y+6=0.5\Rightarrow y_2=-5.5$ $\Rightarrow (36.75,-6.5),(36.75,-5.5)$ Plot the points, draw the directrix, and graph the parabola: