Question

PLEASE HELP!!!!!!!please

Answer 1

Answer:

see explanation

Step-by-step explanation:

To find the zeros let h(t) = 0, that is

t² + 4t + 3 = 0 ← in standard form

(t + 3)(t + 1) = 0 ← in factored form

Equate each factor to zero and solve for t

t + 3 = 0 ⇒ t = - 3 ← smaller t

t + 1 = 0 ⇒ t = - 1 ← larger t

(2)

given a parabola in standard form : ax² + bx + c ( a ≠ 0)

Then the x- coordinate of the vertex is

$x_{vertex}$ = - $\frac{b}{2a}$

h(t) = t² + 4t + 3 ← is in standard form

with a = 1 and b = 4, thus

$x_{vertex}$ = - $\frac{4}{2}$ = - 2

Substitute t = - 2 into h(t) for y- coordinate

h(- 2) = (- 2)² + 4(- 2) + 3 = 4 - 8 + 3 = - 1

Vertex = (- 2, - 1 )