What is the the vertex of #y = x^2-x+16#?

1 Answer
May 11, 2018

#"vertex "=(1/2,63/4)#

Explanation:

#"given a quadratic in standard form"color(white)(x)ax^2+bx+c#

#"then the x-coordinate of the vertex is"#

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

#y=x^2-x+16" is in standard form"#

#"with "a=1,b=-1" and "c=16#

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

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

#y_("vertex")=(1/2)^2-1/2+16=63/4#

#rArrcolor(magenta)"vertex "=(1/2,63/4)#