The given plane, , has normal vector
. Any plane parallel to this one has the same normal vector.
Let be any point in the plane we want. The plane contains the point (1, 1, -1), so an arbitrary vector in this plane is
and this is perpendicular to .
So the equation of the plane is
or equivalently,
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
