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]}
