#"given the equation of a parabola in standard form"#
#•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0#
#"then the x-coordinate of the vertex is"#
#x_(color(red)"vertex")=-b/(2a)#
#y=x^2-2x-6" is in standard form"#
#"with "a=1,b=-2,c=-6#
#rArrx_(color(red)"vertex")=-(-2)/2=1#
#"substitute this value into the equation for y"#
#rArry_(color(red)"vertex")=1-2-6=-7#
#rArrcolor(magenta)"vertex "=(1,-7)#
#color(blue)"for intercepts"#
#• " let x = 0, in equation for y-intercept"#
#• " let y = 0, in equation for x-intercepts"#
#x=0toy=-6larrcolor(red)" y-intercept"#
#y=0tox^2-2x-6=0#
#"solve for x using the "color(blue)"quadratic formula"#
#x=(2+-sqrt(4+24))/2#
#color(white)(x)=(2+-sqrt28)/2#
#color(white)(x)=(2+-2sqrt7)/2=1+-sqrt7#
#rArrx~~ -1.65,x~~ 3.65larrcolor(red)" x-intercepts"#
graph{x^2-2x-6 [-20, 20, -10, 10]}
