What is the length of MN
2 answers:
You might be interested in
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 )