#"the parabola can be graphed by finding the zeros, vertex"#
#"and additional points"#
#"express in standard form "y=ax^2+bx+c#
#rArry=x^2+x#
#"with "a=1,b=1,c=0#
#"since "c=0" then y-intercept is 0"#
#• " if "a>0" then minimum turning point " uuu#
#• " if "a<0" then maximum turning point " nnn#
#"here "a=1>0rArr" minimum turning point"#
#"to find the zeros let y = 0 and solve for x"#
#rArrx^2+x=0#
#rArrx(x+1)=0#
#rArrx=0" or "x=-1larrcolor(red)" zeros"#
#"find the vertex by "color(blue)"completing the square"#
#rArrx^2+2(1/2)x+1/4-1/4#
#=(x+1/2)^2-1/4larrcolor(red)" in vertex form"#
#"the vertex "=(-1/2,-1/4)#
#color(blue)"Additional points"#
#x=1toy=1^2+1=2rArr(1,2)#
#x=2toy=2^2+2=6rArr(2,6)#
#x=-2toy=(-2)^2-2=2rArr(-2,2)#
#"plot these key points and draw a smooth curve through them"#
graph{x^2+x [-10, 10, -5, 5]}
