How do you find the vertex and intercepts for #y=x^2+3x+2#?
1 Answer
Apr 20, 2018
Explanation:
#• " let x = 0 for y-intercept"#
#• " let y = 0 for x-intercepts"#
#x=0rArry=2larrcolor(red)"y-intercept"#
#y=0rArrx^2+3x+2=0#
#"the factors of + 2 which sum to + 3 are + 1 and + 2"#
#rArr(x+1)(x+2)=0#
#"equate each factor to zero and solve for x"#
#x+1=0rArrx=-1#
#x+2=0rArrx=-2#
#rArrx=-2,x=-1larrcolor(red)"are x-intercepts"#
#"the vertex lies on the "color(blue)"axis of symmetry"#
#"which is situated at the midpoint of the intercepts"#
#rArrx_(color(red)"vertex")=(-2-1)/2=-3/2#
#"substitute this value into the equation for y-coordinate"#
#y_(color(red)"vertex")=(-3/2)^2+3(-3/2)+2=-1/4#
#rArrcolor(magenta)"vertex "=(-3/2,-1/4)#
graph{x^2+3x+2 [-10, 10, -5, 5]}