On the graph, plot all relevant points for y=x^2-4x+y^2-2y+8. Then give the equation for a graph that shifts down 1 unit and to the right 2 units. If it is a function, give its domain and range.
Thank you!!
I find it odd that your textbook stated the equation in this unsimplified form, since we could combine the two y terms to read
x^2 - 4x + y^2 - 3y + 8 = 0
From that form I recognize a circle.
let's complete the square:
x^2 - 4x + 4 + y^2 - 3y + 9/4 = -8+4+9/4
(x-2)^2 + (y-3/2)^2 = 1/4
so centre is (2,3/2) and radius is 1/2
we want to move this 1 unit down and 2 to the right, so
(x-2-2)^2 + (y-3/2+1)^2 = 1/4
(x-4)^2 + (y-1/2)^2 = 1/4
expand it if you will
new centre is (4,1/2), radius is still 1/2
It was on a test. I think it was written like that to trick us. Normally I don't have a problem with math, but this teacher is pretty tricky! I really appreciate all of your help=;)
To plot the graph, we need to find the relevant points for the given equation y = x^2 - 4x + y^2 - 2y + 8.
Step 1: Rearrange the terms to get all the variables on one side of the equation:
x^2 - 4x + y^2 - 2y - y - 8 = 0
Step 2: Group the x-terms and the y-terms:
(x^2 - 4x) + (y^2 - 2y - y) - 8 = 0
Step 3: Complete the square separately for x-terms and y-terms:
(x^2 - 4x + 4) + (y^2 - 2y - y + 1) - 8 = 4 + 1
Step 4: Simplify the equation:
(x - 2)^2 + (y - 1)^2 - 8 = 5
Step 5: Rearrange the equation to isolate (x - 2)^2 and (y - 1)^2 terms:
(x - 2)^2 + (y - 1)^2 = 13
Now, we can plot the graph by finding a set of points that satisfy the equation (x - 2)^2 + (y - 1)^2 = 13. For simplicity, we can solve this equation for various values of x and find the corresponding y coordinates.
To find the equation for a graph that shifts down 1 unit and to the right 2 units, we need to alter the equation (x - 2)^2 + (y - 1)^2 = 13 accordingly.
Step 6: Shift the graph down 1 unit and to the right 2 units:
(x - (2 + 2))^2 + (y - (1 + 1))^2 = 13
(x - 4)^2 + (y - 2)^2 = 13
The resulting equation for the shifted graph is (x - 4)^2 + (y - 2)^2 = 13.
Now, let's address whether this graph represents a function and determine its domain and range.
A circle graph (x - 4)^2 + (y - 2)^2 = 13 is not a function because it fails the vertical line test, meaning that there are multiple y-values for a given x-value on the circle.
The domain of this graph includes all real numbers, as there are no restrictions on the x-values. So, the domain is (-∞, +∞).
The range of this graph includes all real numbers within the range of y-values on the circle, which is (-∞, +∞).
In summary:
- The graph consists of points that satisfy (x - 2)^2 + (y - 1)^2 = 13.
- The equation for the shifted graph is (x - 4)^2 + (y - 2)^2 = 13.
- The graph is not a function.
- The domain is (-∞, +∞).
- The range is (-∞, +∞).