What are the vertex, focus and directrix of y=-15+12x-2x^2 ?
1 Answer
Aug 8, 2018
Explanation:
"since the equation has an "x^2" term, this is a"
"vertically opening parabola"
"the equation of a vertically opening parabola is"
•color(white)(x)(x-h)^2=4a(y-k)
"where "(h,k)" are the coordinates of the vertex and a"
"is the distance from the vertex to the focus and directrix"
"if "a>0" then opens upwards"
"if "a< 0" then opens downwards"
"to obtain this form "color(blue)"complete the square"
y=-2(x^2-6x+15/2)
color(white)(y)=-2(x^2+2(-3)x+9-9+15/2)
color(white)(y)=-2(x-3)^2+3
(x-3)^2=-1/2(y-3)
4a=-1/2rArra=-1/8" parabola opens down"
"vertex "=(3,3)
"focus "=(h,a+k)=(3,23/8)
"directrix is "y=-a+k=1/8+3=25/8
graph{(y+2x^2-12x+15)(y-0.001x-25/8)=0 [-10, 10, -5, 5]}