Answer:
Part 1) The function of the First graph is 
Part 2) The function of the Second graph is 
Part 3) The function of the Third graph is
See the attached figure
Step-by-step explanation:
we know that
The quadratic equation in factored form is equal to

where
a is the leading coefficient
c and d are the roots or zeros of the function
Part 1) First graph
we know that
The solutions or zeros of the first graph are
x=-1 and x=3
The parabola open up, so the leading coefficient a is positive
The function is equal to

Find the value of the coefficient a
The vertex is equal to the point (1,-4)
substitute and solve for a



therefore
The function is equal to

Part 2) Second graph
we know that
The solutions or zeros of the first graph are
x=-3 and x=1
The parabola open down, so the leading coefficient a is negative
The function is equal to

Find the value of the coefficient a
The vertex is equal to the point (-1,8)
substitute and solve for a



therefore
The function is equal to

Part 3) Third graph
we know that
The solutions or zeros of the first graph are
x=-2 and x=6
The parabola open up, so the leading coefficient a is positive
The function is equal to

Find the value of the coefficient a
The vertex is equal to the point (2,-8)
substitute and solve for a



therefore
The function is equal to
