How do you find the vertex and intercepts for #y=x^2+4x+1#?

1 Answer
Oct 10, 2017

#(-2,-3),x=-2+-sqrt3#

Explanation:

#"the equation of a parabola in "color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#

#"where "(h,k)" are the coordinates of the vertex and a"#
#"is a multiplier"#

#"to obtain this form use "color(blue)"completing the square"#

#• " ensure the coefficient of "x^2" term is 1"#

#• " add/subtract "(1/2"coefficient of x-term")^2" to "x^2+4x#

#y=x^2+4x+4larr" coefficient of "x^2" term is 1"#

#y=x^2+2(2)xcolor(red)(+4)color(red)(-4)+1#

#color(white)(y)=(x+2)^2-3#

#rArr"vertex "=(-2,-3)#

#color(blue)"Intercepts"#

#• " let x = 0, in the equation for y-intercept"#

#• " let y = 0, in equation for x-intercepts"#

#x=0toy=4-3=1larrcolor(red)" y-intercept"#

#y=0to(x+2)^2-3=0#

#rArr(x+2)^2=3#

#rArrx+2=+-sqrt3larr" note plus or minus"#

#rArrx=-2+-sqrt3larrcolor(blue)"exact values"#

#x~~ -3.73,x~~ -0.27larrcolor(red)" x-intercepts"#