# Write-up 11

### by

## Lynne Bombard

**Problem**: Investigate with different
values of p,

, ,,,

for n>1, n=1, and n<1.

Step I

For the graph of , k=1 the graph
is a parabola that opens to the right and does not matter what the value
of p is. When the absolute value of p increases the graph is wider. The
vertex of this parabola is (-p/2,0). And as we change the value of p, the
y-intercepts are p and -p. When p is negative, the parabola is reflected
over the y-axis.

When |n| >1, the graph is is a hyperbola. A negative
value of p reflects the graph over the y-axis. To obtain the vertices of
the hyperbola, we must evaluate when theta
is pi and 2pi.

When -1<n<1, the graph is an ellipse.

For the graph , its just
a reflection across the y-axis of everything that occurred in step I for
the equation .

For the graph of , k=1 the graph
is a parabola that opens upwards and it does not matter what the value of
p is. When the absolute value of p increases the graph is wider. The vertex
of this parabola is (0,-p/2). And as we change the value of p, the x-intercepts
are p and -p. When p is negative, the parabola is reflected over the x-axis.

When |n| >1, the graph is is a hyperbola. A negative
value of p reflects the graph over the x-axis. To obtain the vertices of
the hyperbola, we must evaluate when theta
is pi/2 and 3pi/2.

When -1<n<1, the graph is an ellipse.

**Return**