What are the vertex, focus and directrix of # y=x^2-x+19 #?

1 Answer
Feb 7, 2018

#"see explanation"#

Explanation:

#"given the equation of a parabola in standard form"#

#•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0#

#"then the x-coordinate of the vertex which is also"#
#"the axis of symmetry is"#

#•color(white)(x)x_(color(red)"vertex")=-b/(2a)#

#y=x^2-x+19" is in standard form"#

#"with "a=1,b=-1" and "c=19#

#rArrx_(color(red)"vertex")=-(-1)/2=1/2#

#"substitute this value into the equation for y"#

#rArry_(color(red)"vertex")=(1/2)^2-1/2+19=75/4#

#rArrcolor(magenta)"vertex "=(1/2,75/4)#

#rArry=(x-1/2)^2+75/4larrcolor(blue)"in vertex form"#

#"the translated form of a vertically opening parabola is"#

#•color(white)(x)(x-h)^2=4p(y-k)#

#"where "(h,k)" are the coordinates of the vertex and"#
#"p is the distance from the vertex to the focus/directrix"#

#rArr(x-1/2)^2=1(y-75/4)larrcolor(blue)"translated form"#

#"with "4p=1rArrp=1/4#

#"the focus lies on the axis of symmetry "x=1/2#

#"since "a>0" then parabola opens up "uuu#

#"hence the focus is "1/4" unit above the vertex and"#
#"the directrix "1/4" unit below the vertex"#

#rArrcolor(magenta)"focus "=(1/2,19)#

#"and equation of directrix is "y=37/2#