To graph the parabola given by the equation \( y = 3x^2 - 6x - 1 \) and plot five points on it, follow these steps:
1. Find the Vertex:
The vertex of a parabola given by the quadratic equation \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex, \( x = -\frac{b}{2a} \). Here, \( a = 3 \), \( b = -6 \), and \( c = -1 \).
- Calculate the x-coordinate of the vertex:
[tex]\[
x = -\frac{-6}{2 \cdot 3} = 1
\][/tex]
- Substitute \( x = 1 \) back into the equation to find the y-coordinate:
[tex]\[
y = 3(1)^2 - 6(1) - 1 = 3 - 6 - 1 = -4
\][/tex]
- The vertex is at \( (1, -4) \).
2. Find Two Points to the Left of the Vertex:
Select \( x_1 \) and \( x_2 \) as 1 unit and 2 units to the left of the vertex \( x \)-coordinate:
- For \( x = 0 \):
[tex]\[
y = 3(0)^2 - 6(0) - 1 = -1
\][/tex]
The point is \( (0, -1) \).
- For \( x = -1 \):
[tex]\[
y = 3(-1)^2 - 6(-1) - 1 = 3 + 6 - 1 = 8
\][/tex]
The point is \( (-1, 8) \).
3. Find Two Points to the Right of the Vertex:
Select \( x_3 \) and \( x_4 \) as 1 unit and 2 units to the right of the vertex \( x \)-coordinate:
- For \( x = 2 \):
[tex]\[
y = 3(2)^2 - 6(2) - 1 = 12 - 12 - 1 = -1
\][/tex]
The point is \( (2, -1) \).
- For \( x = 3 \):
[tex]\[
y = 3(3)^2 - 6(3) - 1 = 27 - 18 - 1 = 8
\][/tex]
The point is \( (3, 8) \).
To summarize, the five points to be plotted on the graph are:
- Vertex: \( (1, -4) \)
- Two points to the left of the vertex: \( (0, -1) \) and \( (-1, 8) \)
- Two points to the right of the vertex: \( (2, -1) \) and \( (3, 8) \)
Now, plot these points on your graph:
1. Plot the vertex \( (1, -4) \).
2. Plot \( (0, -1) \) and \( (-1, 8) \) to the left of the vertex.
3. Plot \( (2, -1) \) and \( (3, 8) \) to the right of the vertex.
4. Draw the curve through these points to sketch the parabola.
