How do you find the vertex of #y=2x^2+10x+8#?
1 Answer
Mar 30, 2018
Explanation:
#"Given the equation of a parabola in "color(blue)"standard form"#
#•color(white)(x)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+8" is in standard form"#
#"with "a=2,b=10" and "c=8#
#rArrx_("vertex")=-10/4=-5/2#
#"substitute this value into the equation for y-coordinate"#
#y_("vertex")=2(-5/2)^2+10(-5/2)+8=-9/2#
#rArrcolor(magenta)"vertex "=(-5/2,-9/2)#
graph{2x^2+10x+8 [-10, 10, -5, 5]}
