Question

I am supposed to find the 4 vertices of the ellipse

1=(((x+1)^2)/4)+(((y-4)^20/9)

How would I go about doing that?

Answer 1

in
1=(((x+1)^2)/4)+(((y-4)^2)/9) (I fixed your typo) or
9(x+1)^2 + 4(y-4)^2 = 36

let x=0
then 9 + 4(y-4)^2 = 36
(y-4)^2 = 27/4
y-4 = ± (3√3)/4

so the y-intercepts are
4 + (3√3)/4 and 4 - (3√3)/4

let y = 0 to find the x-intercepts in the same way

Answer 2

My teacher told me the answers were (-1,7) (-3,4) (1,4) and (-1,1). But i just didn't understand how to do it, the bell rang.

Answer 3

I misread your question.
I found the intercept, but you did not ask for these.

ok

had it been (x^2)/4 + (y^2)/9 = 1
the centre would have been (0,0)

the vertices would have been (2,0), (-2,0), (0,3) and (0,-3)

Your equation has the same shape , but its centre is at (-1,4)
this ellipse has been translated 1 unit to the left, and 4 units up.

so the vertices are

(2,0) ---> (1,4)
(-2,0( ---> (-3,4)
(0,3) ---> (-1,7)
(0,-3) --> (-1,1)
those match your teacher's answers