Properties of Parabolas worksheet with six equations and spaces to identify key features.

To solve the problem, we need to analyze each given equation of a parabola and identify its properties. Let's go through each equation step by step.

1) \( x = -2y^2 - 4y + 70 \)

#### Min/Max value:
- This is a horizontal parabola (since \( x \) is a function of \( y \)).
- The coefficient of \( y^2 \) is negative (\( -2 \)), so the parabola opens to the left.
- The maximum value of \( x \) occurs at the vertex.

#### Vertex:
- The vertex form of a horizontal parabola is \( x = a(y - k)^2 + h \).
- To find the vertex, complete the square:
\[
x = -2(y^2 + 2y) + 70
\]
\[
x = -2\left(y^2 + 2y + 1 - 1\right) + 70
\]
\[
x = -2\left((y + 1)^2 - 1\right) + 70
\]
\[
x = -2(y + 1)^2 + 2 + 70
\]
\[
x = -2(y + 1)^2 + 72
\]
- The vertex is \( (h, k) = (72, -1) \).

#### Axis of Symmetry:
- The axis of symmetry is a vertical line passing through the vertex: \( y = -1 \).

#### Opens:
- Since the coefficient of \( (y + 1)^2 \) is negative, the parabola opens to the left.

#### Latus:
- For a horizontal parabola \( x = a(y - k)^2 + h \), the latus rectum length is \( \frac{1}{|a|} \).
- Here, \( a = -2 \), so the latus rectum length is \( \frac{1}{2} \).

#### y-intercept:
- Set \( x = 0 \):
\[
0 = -2y^2 - 4y + 70
\]
\[
2y^2 + 4y - 70 = 0
\]
\[
y^2 + 2y - 35 = 0
\]
\[
(y + 7)(y - 5) = 0
\]
\[
y = -7 \quad \text{or} \quad y = 5
\]
- y-intercepts are \( (0, -7) \) and \( (0, 5) \).

#### x-intercept:
- Set \( y = 0 \):
\[
x = -2(0)^2 - 4(0) + 70 = 70
\]
- x-intercept is \( (70, 0) \).

Final Answer for 1:

\[
\boxed{
\begin{array}{l}
\text{Min/Max value: Max at } 72 \\
\text{Latus: } \frac{1}{2} \\
\text{y-int: } (0, -7), (0, 5) \\
\text{x-int: } (70, 0) \\
\text{Vertex: } (72, -1) \\
\text{Axis of Symmetry: } y = -1 \\
\text{Opens: Left}
\end{array}
}
\]

2) \( y = 2x^2 - 4x - 6 \)

#### Min/Max value:
- This is a vertical parabola (since \( y \) is a function of \( x \)).
- The coefficient of \( x^2 \) is positive (\( 2 \)), so the parabola opens upwards.
- The minimum value of \( y \) occurs at the vertex.

#### Vertex:
- The vertex form of a vertical parabola is \( y = a(x - h)^2 + k \).
- To find the vertex, complete the square:
\[
y = 2(x^2 - 2x) - 6
\]
\[
y = 2\left(x^2 - 2x + 1 - 1\right) - 6
\]
\[
y = 2\left((x - 1)^2 - 1\right) - 6
\]
\[
y = 2(x - 1)^2 - 2 - 6
\]
\[
y = 2(x - 1)^2 - 8
\]
- The vertex is \( (h, k) = (1, -8) \).

#### Axis of Symmetry:
- The axis of symmetry is a vertical line passing through the vertex: \( x = 1 \).

#### Opens:
- Since the coefficient of \( (x - 1)^2 \) is positive, the parabola opens upwards.

#### Latus:
- For a vertical parabola \( y = a(x - h)^2 + k \), the latus rectum length is \( \frac{1}{|a|} \).
- Here, \( a = 2 \), so the latus rectum length is \( \frac{1}{2} \).

#### y-intercept:
- Set \( x = 0 \):
\[
y = 2(0)^2 - 4(0) - 6 = -6
\]
- y-intercept is \( (0, -6) \).

#### x-intercept:
- Set \( y = 0 \):
\[
0 = 2x^2 - 4x - 6
\]
\[
x^2 - 2x - 3 = 0
\]
\[
(x - 3)(x + 1) = 0
\]
\[
x = 3 \quad \text{or} \quad x = -1
\]
- x-intercepts are \( (3, 0) \) and \( (-1, 0) \).

Final Answer for 2:

\[
\boxed{
\begin{array}{l}
\text{Min/Max value: Min at } -8 \\
\text{Latus: } \frac{1}{2} \\
\text{y-int: } (0, -6) \\
\text{x-int: } (3, 0), (-1, 0) \\
\text{Vertex: } (1, -8) \\
\text{Axis of Symmetry: } x = 1 \\
\text{Opens: Up}
\end{array}
}
\]

3) \( y = -2(x - 3)^2 + 2 \)

#### Min/Max value:
- This is a vertical parabola.
- The coefficient of \( (x - 3)^2 \) is negative (\( -2 \)), so the parabola opens downwards.
- The maximum value of \( y \) occurs at the vertex.

#### Vertex:
- The vertex is directly given in the equation: \( (3, 2) \).

#### Axis of Symmetry:
- The axis of symmetry is a vertical line passing through the vertex: \( x = 3 \).

#### Opens:
- Since the coefficient of \( (x - 3)^2 \) is negative, the parabola opens downwards.

#### Latus:
- For a vertical parabola \( y = a(x - h)^2 + k \), the latus rectum length is \( \frac{1}{|a|} \).
- Here, \( a = -2 \), so the latus rectum length is \( \frac{1}{2} \).

#### y-intercept:
- Set \( x = 0 \):
\[
y = -2(0 - 3)^2 + 2 = -2(9) + 2 = -18 + 2 = -16
\]
- y-intercept is \( (0, -16) \).

#### x-intercept:
- Set \( y = 0 \):
\[
0 = -2(x - 3)^2 + 2
\]
\[
2(x - 3)^2 = 2
\]
\[
(x - 3)^2 = 1
\]
\[
x - 3 = \pm 1
\]
\[
x = 4 \quad \text{or} \quad x = 2
\]
- x-intercepts are \( (4, 0) \) and \( (2, 0) \).

Final Answer for 3:

\[
\boxed{
\begin{array}{l}
\text{Min/Max value: Max at } 2 \\
\text{Latus: } \frac{1}{2} \\
\text{y-int: } (0, -16) \\
\text{x-int: } (4, 0), (2, 0) \\
\text{Vertex: } (3, 2) \\
\text{Axis of Symmetry: } x = 3 \\
\text{Opens: Down}
\end{array}
}
\]

4) \( x = 2(y - 3)^2 - 2 \)

#### Min/Max value:
- This is a horizontal parabola.
- The coefficient of \( (y - 3)^2 \) is positive (\( 2 \)), so the parabola opens to the right.
- The minimum value of \( x \) occurs at the vertex.

#### Vertex:
- The vertex is directly given in the equation: \( (-2, 3) \).

#### Axis of Symmetry:
- The axis of symmetry is a horizontal line passing through the vertex: \( y = 3 \).

#### Opens:
- Since the coefficient of \( (y - 3)^2 \) is positive, the parabola opens to the right.

#### Latus:
- For a horizontal parabola \( x = a(y - k)^2 + h \), the latus rectum length is \( \frac{1}{|a|} \).
- Here, \( a = 2 \), so the latus rectum length is \( \frac{1}{2} \).

#### y-intercept:
- Set \( x = 0 \):
\[
0 = 2(y - 3)^2 - 2
\]
\[
2(y - 3)^2 = 2
\]
\[
(y - 3)^2 = 1
\]
\[
y - 3 = \pm 1
\]
\[
y = 4 \quad \text{or} \quad y = 2
\]
- y-intercepts are \( (0, 4) \) and \( (0, 2) \).

#### x-intercept:
- Set \( y = 0 \):
\[
x = 2(0 - 3)^2 - 2 = 2(9) - 2 = 18 - 2 = 16
\]
- x-intercept is \( (16, 0) \).

Final Answer for 4:

\[
\boxed{
\begin{array}{l}
\text{Min/Max value: Min at } -2 \\
\text{Latus: } \frac{1}{2} \\
\text{y-int: } (0, 4), (0, 2) \\
\text{x-int: } (16, 0) \\
\text{Vertex: } (-2, 3) \\
\text{Axis of Symmetry: } y = 3 \\
\text{Opens: Right}
\end{array}
}
\]

5) \( x = 2y^2 - 24y + 70 \)

#### Min/Max value:
- This is a horizontal parabola.
- The coefficient of \( y^2 \) is positive (\( 2 \)), so the parabola opens to the right.
- The minimum value of \( x \) occurs at the vertex.

#### Vertex:
- To find the vertex, complete the square:
\[
x = 2(y^2 - 12y) + 70
\]
\[
x = 2\left(y^2 - 12y + 36 - 36\right) + 70
\]
\[
x = 2\left((y - 6)^2 - 36\right) + 70
\]
\[
x = 2(y - 6)^2 - 72 + 70
\]
\[
x = 2(y - 6)^2 - 2
\]
- The vertex is \( (h, k) = (-2, 6) \).

#### Axis of Symmetry:
- The axis of symmetry is a vertical line passing through the vertex: \( y = 6 \).

#### Opens:
- Since the coefficient of \( (y - 6)^2 \) is positive, the parabola opens to the right.

#### Latus:
- For a horizontal parabola \( x = a(y - k)^2 + h \), the latus rectum length is \( \frac{1}{|a|} \).
- Here, \( a = 2 \), so the latus rectum length is \( \frac{1}{2} \).

#### y-intercept:
- Set \( x = 0 \):
\[
0 = 2y^2 - 24y + 70
\]
\[
y^2 - 12y + 35 = 0
\]
\[
(y - 7)(y - 5) = 0
\]
\[
y = 7 \quad \text{or} \quad y = 5
\]
- y-intercepts are \( (0, 7) \) and \( (0, 5) \).

#### x-intercept:
- Set \( y = 0 \):
\[
x = 2(0)^2 - 24(0) + 70 = 70
\]
- x-intercept is \( (70, 0) \).

Final Answer for 5:

\[
\boxed{
\begin{array}{l}
\text{Min/Max value: Min at } -2 \\
\text{Latus: } \frac{1}{2} \\
\text{y-int: } (0, 7), (0, 5) \\
\text{x-int: } (70, 0) \\
\text{Vertex: } (-2, 6) \\
\text{Axis of Symmetry: } y = 6 \\
\text{Opens: Right}
\end{array}
}
\]

6) \( y = -2(x - 6)^2 + 2 \)

#### Min/Max value:
- This is a vertical parabola.
- The coefficient of \( (x - 6)^2 \) is negative (\( -2 \)), so the parabola opens downwards.
- The maximum value of \( y \) occurs at the vertex.

#### Vertex:
- The vertex is directly given in the equation: \( (6, 2) \).

#### Axis of Symmetry:
- The axis of symmetry is a vertical line passing through the vertex: \( x = 6 \).

#### Opens:
- Since the coefficient of \( (x - 6)^2 \) is negative, the parabola opens downwards.

#### Latus:
- For a vertical parabola \( y = a(x - h)^2 + k \), the latus rectum length is \( \frac{1}{|a|} \).
- Here, \( a = -2 \), so the latus rectum length is \( \frac{1}{2} \).

#### y-intercept:
- Set \( x = 0 \):
\[
y = -2(0 - 6)^2 + 2 = -2(36) + 2 = -72 + 2 = -70
\]
- y-intercept is \( (0, -70) \).

#### x-intercept:
- Set \( y = 0 \):
\[
0 = -2(x - 6)^2 + 2
\]
\[
2(x - 6)^2 = 2
\]
\[
(x - 6)^2 = 1
\]
\[
x - 6 = \pm 1
\]
\[
x = 7 \quad \text{or} \quad x = 5
\]
- x-intercepts are \( (7, 0) \) and \( (5, 0) \).

Final Answer for 6:

\[
\boxed{
\begin{array}{l}
\text{Min/Max value: Max at } 2 \\
\text{Latus: } \frac{1}{2} \\
\text{y-int: } (0, -70) \\
\text{x-int: } (7, 0), (5, 0) \\
\text{Vertex: } (6, 2) \\
\text{Axis of Symmetry: } x = 6 \\
\text{Opens: Down}
\end{array}
}
\]

Summary of All Answers:

1. \(\boxed{
\begin{array}{l}
\text{Min/Max value: Max at } 72 \\
\text{Latus: } \frac{1}{2} \\
\text{y-int: } (0, -7), (0, 5) \\
\text{x-int: } (70, 0) \\
\text{Vertex: } (72, -1) \\
\text{Axis of Symmetry: } y = -1 \\
\text{Opens: Left}
\end{array}
}\)

2. \(\boxed{
\begin{array}{l}
\text{Min/Max value: Min at } -8 \\
\text{Latus: } \frac{1}{2} \\
\text{y-int: } (0, -6) \\
\text{x-int: } (3, 0), (-1, 0) \\
\text{Vertex: } (1, -8) \\
\text{Axis of Symmetry: } x = 1 \\
\text{Opens: Up}
\end{array}
}\)

3. \(\boxed{
\begin{array}{l}
\text{Min/Max value: Max at } 2 \\
\text{Latus: } \frac{1}{2} \\
\text{y-int: } (0, -16) \\
\text{x-int: } (4, 0), (2, 0) \\
\text{Vertex: } (3, 2) \\
\text{Axis of Symmetry: } x = 3 \\
\text{Opens: Down}
\end{array}
}\)

4. \(\boxed{
\begin{array}{l}
\text{Min/Max value: Min at } -2 \\
\text{Latus: } \frac{1}{2} \\
\text{y-int: } (0, 4), (0, 2) \\
\text{x-int: } (16, 0) \\
\text{Vertex: } (-2, 3) \\
\text{Axis of Symmetry: } y = 3 \\
\text{Opens: Right}
\end{array}
}\)

5. \(\boxed{
\begin{array}{l}
\text{Min/Max value: Min at } -2 \\
\text{Latus: } \frac{1}{2} \\
\text{y-int: } (0, 7), (0, 5) \\
\text{x-int: } (70, 0) \\
\text{Vertex: } (-2, 6) \\
\text{Axis of Symmetry: } y = 6 \\
\text{Opens: Right}
\end{array}
}\)

6. \(\boxed{
\begin{array}{l}
\text{Min/Max value: Max at } 2 \\
\text{Latus: } \frac{1}{2} \\
\text{y-int: } (0, -70) \\
\text{x-int: } (7, 0), (5, 0) \\
\text{Vertex: } (6, 2) \\
\text{Axis of Symmetry: } x = 6 \\
\text{Opens: Down}
\end{array}
}\)