Question

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

Answer 1

`"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]}