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)
y = 0x = y + 2plug 0 in for yx = 0 + 2x = 2
-1.83 . . . . . and . . . . . -1.82
solve for y so if x + y = 0 then y = -x
It is: y = -7 which is a straight line parallel to the x axis
If B=0, then the graph does not depend on the value of y. This is a vertical line at x = C/A
y = 0x = y + 2plug 0 in for yx = 0 + 2x = 2
(0,4) and (-2, 0)
-1.83 . . . . . and . . . . . -1.82
solve for y so if x + y = 0 then y = -x
A straight line through the points (0, 1) and (-0.4, 0).
It is: y = -7 which is a straight line parallel to the x axis
If B=0, then the graph does not depend on the value of y. This is a vertical line at x = C/A
y 2 1 x 0 1 2
The x intercept is at (84, 0) and the y intercept is at (0, 112) and so with a line join the points together which then will form a graph for the given equation.
There are none. For this equation, there is nonreal answer, as the graph of the quadratic does not pass below the x-axis
When (the graph of the equations) the two lines intersect. The equations will tell you what the slopes of the lines are, just look at them. If they are different, then the equations have a unique solution..
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.