1. Homework Answers 1) vertex: (1, 2) focus: (1, 4) Axis of Sym: x = 1 directrix: y = 0 2) Eqn: (y + 3)2 = -12(x – 1) vertex: (1, -3) focus: (-2, -3) Axis of Symm: y = -3 directrix: x = 4
3) Eqn: (x + 4)2 = -4(y – 2)
vertex: (-4, 2)
focus: (-4, 1)
A of S: x = - 4
directrix: y = 3
6) x2 = 8y
7) 4 in

2. PG. 14 HW #19: Write the equation for the given parabola.
Section 7.1 Notes, continued
PG. 14 HW #19: Write the equation for the given parabola.
We can see that the orientation (direction it opens) is:
Think about the parts of the equation we need
for a parabola opening down:
(but p must be negative since it opens down)
We need to find the vertex (h,k) and p.
The vertex is (3, 5).
p always represents the distance from the vertex to the focus,
so p =
Plug h, k, and p into the standard form of the equation:
(x – 3)2 = 4(–1)(y – 5)
and simiplify:
(x – 3)2 = –4(y – 5)
DOWN -1

3. A) focus (2, 1) and vertex (−5, 1)
Example 4: Write an equation for and graph a parabola with the given characteristics.
A) focus (2, 1) and vertex (−5, 1)
PLOT the INFORMATION to determine the orientation:
Think about the parts of the equation we need
for a parabola opening right:
(but p must be positive since it opens right)
We need to find the vertex (h,k) and p.
The vertex is (-5, 1).
p always represents the distance from the vertex to the focus,
so p =
Plug h, k, and p into the standard form of the equation:
(y – 1)2 = 4(7)(x – -5) [be careful about the location of h and k]
and simiplify:
(y – 1)2 = 28(x + 5)
RIGHT 7 V F

4. B) vertex (3, −2), directrix y = −1
Example 4: Write an equation for and graph a parabola with the given characteristics.
B) vertex (3, −2), directrix y = −1 
PLOT the INFORMATION to determine the orientation:
Think about the parts of the equation we need
for a parabola opening down:
(but p must be negative since it opens down)
We need to find the vertex (h,k) and p.
The vertex is (3, -2).
p always represents the distance from the vertex to the focus,
so p =
Plug h, k, and p into the standard form of the equation:
(x – 3)2 = 4(–3)(y – -2)
and simiplify:
(x – 3)2 = –12(y + 2)
DOWN –3 V

5. C) focus (−1, 7), opens up, contains (3, 7)
Example 4: Write an equation for and graph a parabola with the given characteristics.
C) focus (−1, 7), opens up, contains (3, 7) F R
PLOT the INFORMATION
Think about the parts of the equation we need
for a parabola opening up:
(p must be positive since it opens up)
We need to find the vertex (h,k) and p.
The vertex always aligns with the focus, so we know (-1, k).
The h value is -1.
To find the k value, use the ordered pair pattern for the FOCUS on the blue equation sheet.
Focus: (h, k+p) = (-1, 7).
If h = -1, then k+p = 7. Solve for k.
k = 7 – p
This makes the vertex (-1, 7-p).
We now have an h and k that we can plug in, as well as an x-value and y-value (3, 7)
to plug into the standard form.
But which vertex?

6. C) focus (−1, 7), opens up, contains (3, 7)
Example 4: Write an equation for and graph a parabola with the given characteristics.
C) focus (−1, 7), opens up, contains (3, 7) F
What we know:
vertex (-1, 7-p); point on the circle: (3, 7) as (x, y)
Plug h, k, x, and y into standard form of the equation to solve for p.
(3 – -1)2 = 4p(7 – (7-p))
(4)2 = 4p(p)
16 = 4p2
4 = p2
p = 2, but since we know it opens up, p = 2
Since p = 2, find the k-value of the vertex using previous steps:
Focus: (h, k+p) = (-1, 7), where k+p=7, and k = 7 – p
k = 7 – 2 ,
Plug h, k, and p into the standard form of the equation:
(x – -1)2 = 4(2)(y – 5) and simiplify:
k = 5 makes the vertex (-1, 5)
(x + 1)2 = 8( y – 5)