"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]}
