To solve this problem, we need to:

1. Identify the vertex of the parabola.
2. Plot two points on either side of the vertex.

Step 1: Equation of the Parabola

The given equation is: $y = x^2 - 2x - 1$ This is a quadratic equation, and the graph of this equation is a parabola.

Step 2: Vertex of the Parabola

To find the vertex, we use the vertex formula for a quadratic equation $y = ax^2 + bx + c$, where:

• $a = 1$,
• $b = -2$,
• $c = -1$.

The x-coordinate of the vertex is given by: $x = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1$

Now substitute $x = 1$ into the equation to find the y-coordinate: $y = (1)^2 - 2(1) - 1 = 1 - 2 - 1 = -2$ So, the vertex is at $(1, -2)$.

Step 3: Choosing Points on Either Side

We will now choose two points to the left of $x = 1$ (for example $x = 0$ and $x = -1$) and two points to the right (for example $x = 2$ and $x = 3$).

For $x = 0$:

$y = (0)^2 - 2(0) - 1 = -1$ Point: $(0, -1)$

For $x = -1$:

$y = (-1)^2 - 2(-1) - 1 = 1 + 2 - 1 = 2$ Point: $(-1, 2)$

For $x = 2$:

$y = (2)^2 - 2(2) - 1 = 4 - 4 - 1 = -1$ Point: $(2, -1)$

For $x = 3$:

$y = (3)^2 - 2(3) - 1 = 9 - 6 - 1 = 2$ Point: $(3, 2)$

Step 4: Plotting the Points

You can now plot the five points:

1. $(1, -2)$ (the vertex),
2. $(0, -1)$,
3. $(-1, 2)$,
4. $(2, -1)$,
5. $(3, 2)$.

Would you like me to provide more details on graphing this or discuss any part of the solution?

Here are 5 related questions:

1. What is the axis of symmetry for this parabola?
2. How do you determine if a parabola opens upward or downward?
3. What is the role of the coefficient $a$ in a quadratic equation?
4. Can you determine the y-intercept directly from the equation?
5. How would the graph change if $b = 0$ in the equation?

Tip: The vertex is always the minimum or maximum point on the parabola, depending on whether it opens upward or downward.