#"given a parabola in "color(blue)"standard form"#
#•color(white)ax^2+bx+c color(white)(x);a!=0#
#"then the x-coordinate of the vertex which is also the"#
#"axis of symmetry is"#
#•color(white)(x)x_(color(red)"vertex")=-b/(2a)#
#f(x)=2x^2+x-3" is in standard form"#
#"with "a=2, b=1, c=-3#
#rArrx_(color(red)"vertex")=-1/4#
#rArr"axis of symmetry is "x=-1/4#
#"substitute this value into the equation for y-coordinate"#
#y_(color(red)"vertex")=2(-1/4)^2+(-1/4)-3=-25/8#
#rArrcolor(magenta)"vertex "=(-1/4,-25/8)#
#"for "color(blue)"intercepts"#
#• " let x = 0, in the equation for y-intercept"#
#• " let y = 0, in the equation for x-intercepts"#
#x=0toy=-3larrcolor(red)"y-intercept"#
#y=0to2x^2+x-3=0#
#rArr(x-1)(2x+3)=0#
#rArrx=1" and "x=-3/2larrcolor(red)"x-intercepts"#
graph{(y-2x^2-x+3)(y-1000x-250)=0 [-10, 10, -5, 5]}
