How do you find the vertex and the intercepts for #f(x)=(x+3)^2+1#?

1 Answer
Aug 6, 2018

#"vertex "=(-3,1)," no x-intercepts"#

Explanation:

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

#•color(white)(x)y=a(x-h)^2+k#

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

#y=(x+3)^2+1" is in this form"#

#color(magenta)"vertex "=(-3,1)#

#"let x=0, for y-intercept"#

#y=9+1=10larrcolor(red)"y-intercept"#

#"let y = 0, for x-intercepts"#

#(x+3)^2+1=0#

#(x+3)^2=-1#

#"this has no real solutions hence there are no"#
#"x-intercepts"#
graph{(x+3)^2+1 [-20, 20, -10, 10]}