What is the expanded form for 1.67822
1 answer:
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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
= -
h(t) = t² + 4t + 3 ← is in standard form
with a = 1 and b = 4, thus
= -
= - 2
Substitute t = - 2 into h(t) for y- coordinate
h(- 2) = (- 2)² + 4(- 2) + 3 = 4 - 8 + 3 = - 1
Vertex = (- 2, - 1 )
I assume the equation is
then let .
The right side is real-valued and positive, and only on the left side do we have an imaginary term, which tells us
Then
but again we omit the negative solution, so that