OK, so the graph is a parabola, with points x=0,y=0; x=6,y=-9; and x=12,y=0
Because the roots of the equation are 0 and 12, we know the formula is therefore of the form
y = ax(x - 12), for some a
So put in x = 6
-9 = 6a(-6)
9 = 36a
a = 1/4
So the parabola has a curve y = x(x-12) / 4, which can also be written y = 0.25x² - 3x
The gradient of this is dy/dx = 0.5x - 3
The key property of a parabolic dish is that it focuses radio waves travelling parallel to the y axis to a single point. So we should arrive at the same focal point no matter what point we chose to look at. So we can pick any point we like - e.g. the point x = 4, y = -8
Gradient of the parabolic mirror at x = 4 is -1
So the gradient of the normal to the mirror at x = 4 is therefore 1.
Radio waves initially travelling vertically downwards are reflected about the normal - which has a gradient of 1, so they're reflected so that they are travelling horizontally. So they arrive parallel to the y axis, and leave parallel to the x axis.
So the focal point is at y = -8, i.e. 1 metre above the back of the dish.
Answer: angle 2 is supplementary to angle 1
i have know clue what it is
It is already in radical form......... but, if you want us to make it into exponential form, it is 8x^(2/3)
Answer:
Remember, 
and the range of g must be in the domain of f.
a)
The domain of f(g(x)) and g(f(x)) is the set of reals.
b)
The domain of f(g(x)) is the set of nonnegative reals and the domain of g(f(x)) is the set of number such that 
c)
The domain of f(g(x)) is the set of reals except the 1 and the domain of g(f(x)) is the set of reals except the 1 and -1
d)
The domain of f(g(x)) is the set of reals except 2, and the domain of g(f(x)) is the set of reals except -1.
e)
The domain of f(g(x)) is the set of nonnegative reals except -3. The domain of g(f(x)) is the set of nonnegative reals except -2.
