How do you find the vertex of #y=2x^2-10x+3#?

1 Answer
Aug 3, 2018

#"vertex "=(5/2,-19/2)#

Explanation:

#"given a parabola in standard form"#

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

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

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

#y=2x^2-10x+3" is in standard form"#

#"with "a=2,b=-10" and "c=3#

#x_("vertex")=-(-10)/4=5/2#

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

#y_("vertex")=2(5/2)^2-10(5/2)+3=-19/2#

#color(magenta)"vertex "=(5/2,-19/2)#
graph{2x^2-10x+3 [-20, 20, -10, 10]}