How do you Find the vertex, Focus, and directrix of the Parabola and sketch its graph #(x-3) + (y-2)^2=0#?

1 Answer
Aug 7, 2018

#"see explanation"#

Explanation:

#"the equation of a horizontally opening parabola is"#

#•color(white)(x)(y-k)^2=4a(x-h)#

#"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" opens to the right "#

#"If "a<0" opens to the left"#

#"rearrange the given equation into this form"#

#(y-2)^2=-(x-3)#

#"with vertex "=(3,2)#

#4a=-1rArra=-1/4" so opens to the left"#

#"Focus "=(a+h,k)=(-1/4+3,2)=(11/4,2)#

#"directrix is "x=-a+h=1/4+3=13/4#

#"for x-intercept let y = 0"#

#4=-x+3rArrx=-1#

#"for y-intercepts let x = 0"#

#(y-2)^2=3rArr(y-2)=+-sqrt3#

#y=2+-sqrt3#

#y~~3.73,y~~0.27#

#"Plot the above coordinates for vertex, focus and "#
#"intercepts and draw a smooth curve through them"#
graph{(y-2)^2=-(x-3) [-10, 10, -5, 5]}