Sketch and determine the turning points of the curve y=x^2-x+1?

1 Answer
Sep 5, 2017

#"minimum turning point at "(1/2,3/4)#

Explanation:

#"this is a quadratic function which can be sketched in "#
#"the usual way"#

#y=x^2-x+1#

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

#"since "a>0" then minimum turning point "uuu#

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

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

#rArry_(color(red)"vertex")=(1/2)^2-1/2+1=3/4#

#rArr"minimum turning point at "(1/2,3/4)#

#"the discriminant shows"#

#Delta=b^2-4ac=1-4=-3#

#"that the quadratic does not intersect the x-axis"#
graph{x^2-x+1 [-10, 10, -5, 5]}

#color(blue)"turning point using calculus"#

#y=x^2-x+1#

#rArrdy/dx=2x-1#

#"for turning point equate "dy/dx=0#

#rArr2x-1=0rArrx=1/2#

#rArr"turning point at "(1/2,3/4)#

#"to test if max/min use the "color(blue)"second derivative test"#

#• " if "(d^2y)/(dx^2)" at "x=1/2>0" then minimum"#

#• " if "(d^2y)/(dx^2)" at "x=1/2<0" then maximum"#

#dy/dx=2x-1#

#rArr(d^2y)/(dx^2)=2>0#

#rArr"minimum turning point at "(1/2,3/4)#