Answer:
\(f(t)=(t+2)^2 -18\) (vertex form)
Vertex(-2,-18), minimum
axis of symmetry at x= -2
Step-by-step explanation:
f(t) = t^2+ 4t − 14
(a) Apply completing the square method
Take middle term +4, divide it by 2 and then square it
4/2= 2 , (2)^2 = 4
add and subtract 4
f(t) = (t^2+ 4t +4) -4− 14
\(f(t)=(t+2)^2 -18\) (vertex form)
(b) Vertex form is y=a(x-h)^2 + k
vertex is (h,k)
when 'a' is positive then vertex is at minimum
when 'a' is negative then vertex is maximum
\(f(t)=(t+2)^2 -18\)
a=1, h=-2 and k= -18
vertex is (-2,-18) , a= 1 that is positive
so it is a minimum
(c) Axis of symmetry at x=h
so axis of symmetry at x= -2
