This rearrangement "shows"
that the vertex is at (h,
k) = (0, 0). The axis of symmetry
is the vertical line right through the vertex: x
= 0. (I can always check my graph,
if I'm not sure about this.) The focus is "p"
units from the vertex. Since the focus is "inside" the parabola
and since this is a "right side up" graph, the focus has to
be above the vertex.

From the conics form of the equation,
shown above, I look at what's multiplied on the unsquared part
and see that 4p
= 4, so p
= 1. Then the focus is one unit above
the vertex, at (0, 1), and
the directrix is the horizontal line y
= –1, one unit below the vertex.

vertex: (0,
0); focus:
(0,
1); axis
of symmetry: x
= 0; directrix:
y
= –1

Graph y^{2}
+ 10y + x + 25 = 0,
and state the vertex, focus, axis of symmetry, and directrix.

Since the y
is squared in this equation, rather than the x,
then this is a "sideways" parabola. To graph, I'll do my T-chart
backwards, picking y-values
first and then finding the corresponding x-values
for x = –y^{2}
– 10y – 25:

To convert the equation into conics form
and find the exact vertex, etc, I'll need to convert the equation to
perfect-square form. In this case, the squared side is already a perfect
square, so:

y^{2} + 10y
+ 25 = –x (y
+ 5)^{2} = –1(x – 0)

This tells me that 4p
= –1, so p
= –1/4. Since the parabola opens
to the left, then the focus is 1/4
units to the left of the vertex. I can see from the equation above that
the vertex is at (h,
k) = (0, –5), so then the
focus must be at (–1/4,
–5). The parabola is sideways, so
the axis of symmetry is, too. The directrix, being perpendicular to
the axis of symmetry, is then vertical, and is 1/4
units to the right of the vertex. Putting this all together, I get:

vertex: (0,
–5); focus:
(–1/4,
–5); axis
of symmetry: y
= –5; directrix:
x
= 1/4

Find the vertex and
focus of y^{2}
+ 6y + 12x – 15 = 0

The y part
is squared, so this is a sideways parabola. I'll get the y stuff
by itself on one side of the equation, and then complete the square
to convert this to conics form.

Then the vertex is at (h,
k) = (2, –3) and the value
of p
is –3.
Since y is
squared and p is
negative, then this is a sideways parabola that opens to the left. This
puts the focus 3 units
to the left of the vertex.

Stapel, Elizabeth.
"Conics: Parabolas: Finding Information From the Equation."
Purplemath. Available fromhttp://www.purplemath.com/modules/parabola2.htm.
Accessed