It is the point of origin of the x and y axes of the graph
There needs to be an x variable to create a graph.
solve for y so if x + y = 0 then y = -x
First convert it to Y= y-x+4=0 y=x-4 The graph has a slope of 1 and the y-intercept is (0,-4)
0 5 /--------------/-----•------/
It is the point of origin of the x and y axes of the graph
There needs to be an x variable to create a graph.
You may mean, what is the graph of the function y = x^2 + 3. This graph shows a upward parabola with a y-intercept of 3 and a minimum at x=0.
solve for y so if x + y = 0 then y = -x
First convert it to Y= y-x+4=0 y=x-4 The graph has a slope of 1 and the y-intercept is (0,-4)
0 5 /--------------/-----•------/
First, reflect the graph of y = x² in the x-axis (line y = 0) to obtain the graph of y = -x²; then second, shift it 3 units up to obtain the graph of y = -x² + 3.
If you mean: y = -15x+60 then 15x+y = 60 When y = 0 then x = 4 When x = 0 then y = 60 Coordinates of the line are: (4, 0) and (0, 60)
Just one. It's at the origin. (0, 0)
Simple factoring. X^2 + 4.5X = 0 factor out X X(X + 4.5) = 0 X = 0 X = -4.5 graph on TI-84 agrees
The line 8x-8=0 is the same as x=1. The graph looks like a vertical line that intersects the x axis at x=1.
When y is 0 then x intercept is at (10, 0) When x is 0 then y intercept is at (0, -12)