How do you find the vertex and intercepts for #y=x^2+3x+2#?

1 Answer
Apr 20, 2018

#(-3/2,-1/4)" and "x=-2,x=-1#

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]}