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To solve the problem of graphing the quadratic functions, we need to follow a systematic approach for each function. Here's how we can do it step by step:

General Steps for Graphing Quadratic Functions:

1. Identify the type of parabola (opening up or down) based on the coefficient of \( x^2 \).
2. Find the vertex using the formula \( x = -\frac{b}{2a} \) for the standard form \( y = ax^2 + bx + c \).
3. Calculate the y-coordinate of the vertex by substituting \( x \) back into the equation.
4. Find the y-intercept by setting \( x = 0 \).
5. Find the x-intercepts (if any) by setting \( y = 0 \) and solving for \( x \).
6. Plot additional points if necessary to get a better shape of the parabola.
7. Sketch the parabola using the vertex, intercepts, and additional points.

Let's go through each function one by one.

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1. \( y = 2x^2 - 1 \)

- Type of parabola: Opens up (since \( a = 2 > 0 \)).
- Vertex: The vertex form is \( y = a(x-h)^2 + k \). Here, it is already in the form \( y = 2x^2 - 1 \), so the vertex is \( (0, -1) \).
- y-intercept: When \( x = 0 \), \( y = -1 \). So, the y-intercept is \( (0, -1) \).
- x-intercepts: Set \( y = 0 \):
\[
2x^2 - 1 = 0 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}
\]
So, the x-intercepts are \( \left( \frac{\sqrt{2}}{2}, 0 \right) \) and \( \left( -\frac{\sqrt{2}}{2}, 0 \right) \).

- Additional points: Choose \( x = 1 \) and \( x = -1 \):
\[
y(1) = 2(1)^2 - 1 = 1 \quad \text{and} \quad y(-1) = 2(-1)^2 - 1 = 1
\]
So, the points are \( (1, 1) \) and \( (-1, 1) \).

- Graph: Plot the vertex \( (0, -1) \), y-intercept \( (0, -1) \), x-intercepts \( \left( \frac{\sqrt{2}}{2}, 0 \right) \) and \( \left( -\frac{\sqrt{2}}{2}, 0 \right) \), and additional points \( (1, 1) \) and \( (-1, 1) \). Sketch the parabola.

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2. \( y = -2x^2 - 4x + 2 \)

- Type of parabola: Opens down (since \( a = -2 < 0 \)).
- Vertex: Use \( x = -\frac{b}{2a} \):
\[
x = -\frac{-4}{2(-2)} = -\frac{4}{-4} = 1
\]
Substitute \( x = 1 \) into the equation:
\[
y = -2(1)^2 - 4(1) + 2 = -2 - 4 + 2 = -4
\]
So, the vertex is \( (1, -4) \).
- y-intercept: When \( x = 0 \), \( y = 2 \). So, the y-intercept is \( (0, 2) \).
- x-intercepts: Set \( y = 0 \):
\[
-2x^2 - 4x + 2 = 0 \implies x^2 + 2x - 1 = 0
\]
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
x = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}
\]
So, the x-intercepts are \( (-1 + \sqrt{2}, 0) \) and \( (-1 - \sqrt{2}, 0) \).

- Additional points: Choose \( x = -1 \) and \( x = 2 \):
\[
y(-1) = -2(-1)^2 - 4(-1) + 2 = -2 + 4 + 2 = 4 \quad \text{and} \quad y(2) = -2(2)^2 - 4(2) + 2 = -8 - 8 + 2 = -14
\]
So, the points are \( (-1, 4) \) and \( (2, -14) \).

- Graph: Plot the vertex \( (1, -4) \), y-intercept \( (0, 2) \), x-intercepts \( (-1 + \sqrt{2}, 0) \) and \( (-1 - \sqrt{2}, 0) \), and additional points \( (-1, 4) \) and \( (2, -14) \). Sketch the parabola.

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3. \( y = -3x^2 - 6x - 1 \)

- Type of parabola: Opens down (since \( a = -3 < 0 \)).
- Vertex: Use \( x = -\frac{b}{2a} \):
\[
x = -\frac{-6}{2(-3)} = -\frac{6}{-6} = -1
\]
Substitute \( x = -1 \) into the equation:
\[
y = -3(-1)^2 - 6(-1) - 1 = -3 + 6 - 1 = 2
\]
So, the vertex is \( (-1, 2) \).
- y-intercept: When \( x = 0 \), \( y = -1 \). So, the y-intercept is \( (0, -1) \).
- x-intercepts: Set \( y = 0 \):
\[
-3x^2 - 6x - 1 = 0
\]
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-3)(-1)}}{2(-3)} = \frac{6 \pm \sqrt{36 - 12}}{-6} = \frac{6 \pm \sqrt{24}}{-6} = \frac{6 \pm 2\sqrt{6}}{-6} = -1 \mp \frac{\sqrt{6}}{3}
\]
So, the x-intercepts are \( \left( -1 + \frac{\sqrt{6}}{3}, 0 \right) \) and \( \left( -1 - \frac{\sqrt{6}}{3}, 0 \right) \).

- Additional points: Choose \( x = -2 \) and \( x = 1 \):
\[
y(-2) = -3(-2)^2 - 6(-2) - 1 = -12 + 12 - 1 = -1 \quad \text{and} \quad y(1) = -3(1)^2 - 6(1) - 1 = -3 - 6 - 1 = -10
\]
So, the points are \( (-2, -1) \) and \( (1, -10) \).

- Graph: Plot the vertex \( (-1, 2) \), y-intercept \( (0, -1) \), x-intercepts \( \left( -1 + \frac{\sqrt{6}}{3}, 0 \right) \) and \( \left( -1 - \frac{\sqrt{6}}{3}, 0 \right) \), and additional points \( (-2, -1) \) and \( (1, -10) \). Sketch the parabola.

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4. \( y = x^2 + 6x \)

- Type of parabola: Opens up (since \( a = 1 > 0 \)).
- Vertex: Use \( x = -\frac{b}{2a} \):
\[
x = -\frac{6}{2(1)} = -3
\]
Substitute \( x = -3 \) into the equation:
\[
y = (-3)^2 + 6(-3) = 9 - 18 = -9
\]
So, the vertex is \( (-3, -9) \).
- y-intercept: When \( x = 0 \), \( y = 0 \). So, the y-intercept is \( (0, 0) \).
- x-intercepts: Set \( y = 0 \):
\[
x^2 + 6x = 0 \implies x(x + 6) = 0 \implies x = 0 \text{ or } x = -6
\]
So, the x-intercepts are \( (0, 0) \) and \( (-6, 0) \).

- Additional points: Choose \( x = -4 \) and \( x = -2 \):
\[
y(-4) = (-4)^2 + 6(-4) = 16 - 24 = -8 \quad \text{and} \quad y(-2) = (-2)^2 + 6(-2) = 4 - 12 = -8
\]
So, the points are \( (-4, -8) \) and \( (-2, -8) \).

- Graph: Plot the vertex \( (-3, -9) \), y-intercept \( (0, 0) \), x-intercepts \( (0, 0) \) and \( (-6, 0) \), and additional points \( (-4, -8) \) and \( (-2, -8) \). Sketch the parabola.

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

- Type of parabola: Opens down (since \( a = -1 < 0 \)).
- Vertex: Use \( x = -\frac{b}{2a} \):
\[
x = -\frac{-6}{2(-1)} = -\frac{6}{-2} = -3
\]
Substitute \( x = -3 \) into the equation:
\[
y = -(-3)^2 - 6(-3) - 4 = -9 + 18 - 4 = 5
\]
So, the vertex is \( (-3, 5) \).
- y-intercept: When \( x = 0 \), \( y = -4 \). So, the y-intercept is \( (0, -4) \).
- x-intercepts: Set \( y = 0 \):
\[
-x^2 - 6x - 4 = 0 \implies x^2 + 6x + 4 = 0
\]
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
x = \frac{-6 \pm \sqrt{36 - 16}}{2} = \frac{-6 \pm \sqrt{20}}{2} = \frac{-6 \pm 2\sqrt{5}}{2} = -3 \pm \sqrt{5}
\]
So, the x-intercepts are \( (-3 + \sqrt{5}, 0) \) and \( (-3 - \sqrt{5}, 0) \).

- Additional points: Choose \( x = -4 \) and \( x = -2 \):
\[
y(-4) = -(-4)^2 - 6(-4) - 4 = -16 + 24 - 4 = 4 \quad \text{and} \quad y(-2) = -(-2)^2 - 6(-2) - 4 = -4 + 12 - 4 = 4
\]
So, the points are \( (-4, 4) \) and \( (-2, 4) \).

- Graph: Plot the vertex \( (-3, 5) \), y-intercept \( (0, -4) \), x-intercepts \( (-3 + \sqrt{5}, 0) \) and \( (-3 - \sqrt{5}, 0) \), and additional points \( (-4, 4) \) and \( (-2, 4) \). Sketch the parabola.

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6. \( y = -(x-5)^2 - 3 \)

- Type of parabola: Opens down (since the coefficient of \( (x-5)^2 \) is negative).
- Vertex: The vertex form is \( y = a(x-h)^2 + k \). Here, the vertex is \( (5, -3) \).
- y-intercept: When \( x = 0 \):
\[
y = -(0-5)^2 - 3 = -25 - 3 = -28
\]
So, the y-intercept is \( (0, -28) \).
- x-intercepts: Set \( y = 0 \):
\[
-(x-5)^2 - 3 = 0 \implies -(x-5)^2 = 3 \implies (x-5)^2 = -3
\]
Since \( (x-5)^2 \) cannot be negative, there are no real x-intercepts.

- Additional points: Choose \( x = 4 \) and \( x = 6 \):
\[
y(4) = -(4-5)^2 - 3 = -(-1)^2 - 3 = -1 - 3 = -4 \quad \text{and} \quad y(6) = -(6-5)^2 - 3 = -(1)^2 - 3 = -1 - 3 = -4
\]
So, the points are \( (4, -4) \) and \( (6, -4) \).

- Graph: Plot the vertex \( (5, -3) \), y-intercept \( (0, -28) \), and additional points \( (4, -4) \) and \( (6, -4) \). Sketch the parabola.

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Final Answer:

\[
\boxed{
\begin{array}{c}
\text{Graph each quadratic function as described above.}
\end{array}
}
\]