What is the equation of the following graph in vertex form?
y = (x − 3)^2 − 1
y = (x + 3)^2 − 1
y = (x − 4)^2 − 2
y = (x − 4)^2 + 8

Choice B.   y = (x+3)^2 - 1

How I got that

The vertex is the lowest point of parabolas that open upward.

(h,k) = vertex

(h,k) = (-3,-1)

h = -3

k = -1

For each of the answer choices, a = 1.

The general template of a quadratic in vertex form is

y = a(x-h)^2 + k

Plug a = 1, h = -3, k = -1 into that equation. Simplify.

y = a(x-h)^2 + k

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

y = (x+3)^2 - 1

Ibelieve the answer is c

Answer to your question: a. the fundamental theorem of algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. the roots are ​ x=1±i7√2 .number of roots=degree of polynomialhere, degree of polynomial is 2 so number of roots will be 2.hope it *_*