What is the vertex of # y= 6(x-3)^2-x^2-5x+3#?
1 Answer
Aug 12, 2018
Explanation:
#"expand and express in standard form"#
#y=6(x^2-6x+9)-x^2-5x+3#
#color(white)(x)=6x^2-36x+54-x^2-5x+3#
#color(white)(y)=5x^2-41x+57larrcolor(blue)"in standard form"#
#"with "a=5,b=-41" and "c=57#
#"given the equation in standard form then the x-coordinate"#
#"of the vertex is"#
#•color(white)(x)x_(color(red)"vertex")=-b/(2a)#
#x_("vertex")=-(-41)/10=41/10#
#"substitute this value into the equation for y-coordinate"#
#y_("vertex")=5(41/10)^2-41(41/10)+57#
#color(white)(xxxx)=1681/20-3362/20+1140/20=-541/20#
#color(magenta)"vertex "=(41/10,-541/20)#
