I need help with the following questions.

Graph the parabola with the given equation.

(y-2)^2=6(x-1)

(x+1)^2=6(y-1)

Factor each one out...

2y-4 = 6x-6

2x+2 = 6y-6

Factoring doesn't help much.

An easy method of graphing parabolas is to make a T-chart (with x and y). Pick a few x-values (like -2, -1, 0, 1, and 2) and then calculate the y-values. I wouldn't really recommend this.

Learning the rules for graphing parabolas will be much more useful. I would give you the link to a website with some good information, but I'm not allowed to post links. You can google "graph parabola" without quotes to find some helpful information.

To graph the given parabolas, you can follow these steps:

1. Identify the vertex: The vertex form of a parabola is given as (h, k), where "h" represents the x-coordinate and "k" represents the y-coordinate of the vertex.

For the first equation, (y - 2)^2 = 6(x - 1), the vertex is (1, 2).
For the second equation, (x + 1)^2 = 6(y - 1), the vertex is (-1, 1).

2. Determine the axis of symmetry: The axis of symmetry is a vertical line that passes through the vertex. It is given by the equation x = h, where "h" is the x-coordinate of the vertex.

For the first equation, the axis of symmetry is x = 1.
For the second equation, the axis of symmetry is y = -1.

3. Calculate additional points: To plot more points on the graph, you can substitute various values for either x or y and solve for the other variable. It is helpful to choose values that are on either side of the vertex.

For the first equation, you can choose different values of x, substitute them into the equation, and solve for y. For example:

- When x = 0: (y - 2)^2 = 6(0 - 1) => y - 2 = -6 => y = -4.
So one point on the graph is (0, -4).
- When x = 2: (y - 2)^2 = 6(2 - 1) => y - 2 = 6 => y = 8.
So another point on the graph is (2, 8).

For the second equation, you can choose different values of y, substitute them into the equation, and solve for x. For example:

- When y = -2: (x + 1)^2 = 6(-2 - 1) => (x + 1)^2 = -18. This equation has no real solutions, indicating that there are no real points on the graph for this particular value of y.
- When y = 0: (x + 1)^2 = 6(0 - 1) => (x + 1)^2 = -6. Again, this equation has no real solutions.

4. Plot the points: Use the vertex and the additional points you calculated to plot the graph on a coordinate plane. Be sure to label the axis of symmetry.

For the first equation, the vertex is (1, 2), and additional points are (0, -4) and (2, 8). To graph this equation, plot the points and sketch a smooth curve that passes through them.

For the second equation, the vertex is (-1, 1), but since there were no real solutions for any other y-values, the graph will only consist of the vertex plotted.

Note: If you have access to graphing software or a graphing calculator, it can make the plotting process easier and provide a more accurate representation of the parabola.