Is the section 9.1 going to be on our 2nd test?
No. The only topic on the test is in the chapters on trigonometry.
how do you find the equation of a parabola satisfying the given conditions with a focus of (4,0), and a directrix x= -4 ?
The equation of a parabola opening in the x direction is x = 1 / (4 a) * y^2, where a is the displacement from vertex to focus or from directrix to vertex.
Since the vertex is halfway between focus and directrix, it follows that the displacement from vertex to focus or directrix to vertex is half the displacement from directrix to focus.
From directrix at x = -4 to focus at (4, 0) the displacement is +8 units. So a is half of this, or 4.
The equation is therefore
x = 1 / (4 * 4) * y^2 or
x = 1/16 * y^2, or
x = y^2 / 16.
Is the parabola always the graph of a function?
No. A parabola with vertical axis of symmetry is always a shifted and stretched version of y = x^2.
However a 'vertical' parabola is rotated so that its axis is no longer vertical it will no longer represent the graph of a function.
The parabola in the above example does not represent the graph of a function, since it lies 'on its side' (its axis is horizontal). This parabola clearly fails the vertical line test.