You have this equation:
x2 - 12y = 0
First, move the term with y to the right side using addition:
x2 = 12y
Now, divide both sides by 12, so y is by itself:
x2/12 = y
From here, you can use the properties of a parabola (y = x2), "flattening" it so that it is 1/12 of the usual height. You can start with the fact that a parabola is symmetrical over the y-axis-- the vertical line through the point (0,0)-- and passes through the point (0,0). From that point, it gradually turns upward as it moves away from (0,0), passing through the points (1, 1/12), (2, 4/12), (3, 9/12), and so on.
Just one. It's at the origin. (0, 0)
x intercept = (5, 0) y intercept = (0, -2)
You draw a straight line through (0,b) which has a slope of m.
The graph of the function f(x) = 4, is the horizontal line to the x=axis, which passes through (0, 4). The domain of f is all real numbers, and the range is 4.
x=y+2 y=x-2 The y value at the x axis (x=0) will be -2, so graph (0, -2). Let's calculate a few more points by varying x and calculating y: if x=2, y=2-2=0 (2, 0) similarly: (1, -1) (5, 3) Graph those points, then draw a line connecting them all. That's the graph of x=y+2.
The highest point on a graph is when the derivative of the graph equals 0 or the slope is constant.
The equation 0 equals 0 is an identity and contributes absolutely nothing to the part of the graph that you should shade or not. The tautological statement can be ignored.
There needs to be an x variable to create a graph.
there is no right answer
0 5 /--------------/-----•------/
Just one. It's at the origin. (0, 0)
When x = 0, the point that has (0, y) coordinates will be on the y-axis for any y.
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)
It is a horizontal line (slope of 0) which intersects the coordinate (0,5).
a straight line (the Y-axis of a coordinate plane)
the graph is 2 straight lines from the origin in quadrants 1 and 2 at angle of 45 and 135 degrees from 0
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.