It is the y-axis, or the yz plane.
The x-intercept of a graph is the point where the y-coordinate is 0. It represents the value of x at which the graph intersects the x-axis. To find the x-intercept, you can set the equation of the graph equal to zero and solve for x.
It seems there may be a typo in your function notation. If you meant to express the function as ( f(x) = -2(x^3) + 5 ), you can find a point on the graph by choosing a value for ( x ). For example, if ( x = 0 ), then ( f(0) = -2(0^3) + 5 = 5 ). Thus, the point ( (0, 5) ) lies on the graph of the function.
It is the point of origin of the x and y axes of the graph
The lowest point is y = 0 when x = 0. All other values of y are greater than zero for any other value of x
It seems there might be a small error in your expression for the function f(x). If you meant f(x) = (x - 4)^2, then any point on the graph can be found by substituting a value for x. For example, if you plug in x = 4, f(4) = (4 - 4)^2 = 0, so the point (4, 0) lies on the graph. If you clarify the function, I can provide more specific points.
It seems there may be a typo in your function notation. If you meant to express the function as ( f(x) = -2(x^3) + 5 ), you can find a point on the graph by choosing a value for ( x ). For example, if ( x = 0 ), then ( f(0) = -2(0^3) + 5 = 5 ). Thus, the point ( (0, 5) ) lies on the graph of the function.
It is the point of origin of the x and y axes of the graph
A point can represent a piece of data or an (x,y) value.
point
It is the x intercept
The lowest point is y = 0 when x = 0. All other values of y are greater than zero for any other value of x
x=0
point
point
I am assuming the you are talking about the graph of the derivative. The graph of the derivative of F(x) is the graph such that, for any x, the value of x on the graph of the derivative of F(x) is the slope at point x in F(x).
If the discriminant = 0 then the graph touches the x axis at one point If the discriminant > 0 then the graph touches the x axis at two ponits If the discriminant < 0 then the graph does not meet the x axis
x -3y = 0 -x = -x -3y=-x /-3 = /-3 y=1/3x Then solve y for different values of x, record the data , then graph the x and y position for each value of x. so for if x =1 y = 1/3 so one point on the graph is (1,1/3)