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

1 Answer
Aug 6, 2018

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

Explanation:

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

•color(white)(x)y=a(x-h)^2+kxy=a(xh)2+k

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

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

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

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

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

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

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

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

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