(x-h)^2=4P(y-k), vertex is (h,k) P is distance from vertex to directix remember to subtract P from the y value of the vertex (p-k) and that y value is the directix, y=p-k
nut ok so one way is to just graph them on a graphing utility remember if the graph opens up, then the directix is below that or we can convert to 4P(y-k)=(x-h)^2 form where P is distance from directix
I will only convert the 1st one fully, you should be able to do the rest
1. y=-x^2+3x+8 multiply both sides by -1 since we don't like the x^2 term negative -y=x^2-3x-8 add8 to both sides -y+8=x^2-3x take 1/2 of linear coeficient and square it and add to both sides -3/2=-1.5 (-1.5)^2=2.25 -y+10.25=x^2-3x+2.25 factor perfect square -y+10.25=(x-1.5)^2 force undistribute -1 in left side (-1)(y-10.25)=something, we don't care anymore for now factor out a 4 in -1 4(-1/4)(y-10.25)
k=10.25 p=-1/4=-0.25 directix=k-p=10.25-(-0.25)=10.5 directix is y=10.5
basically completee the square with x and find P by force factoring a 4 out
