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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)
It is the point of origin of the x and y axes of the graph
a constant y=x-3 dy/dx=0 a straight line
5
-3
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)
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.
Move 3 over the right side of the equation so the equation would be x = -3. The graph of this would be a verticle line at x= -3
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.
The Equation of a Rational Function has the Form,... f(x) = g(x)/h(x) where h(x) is not equal to zero. We will use a given Rational Function as an Example to graph showing the Vertical and Horizontal Asymptotes, and also the Hole in the Graph of that Function, if they exist. Let the Rational Function be,... f(x) = (x-2)/(x² - 5x + 6). f(x) = (x-2)/[(x-2)(x-3)]. Now if the Denominator (x-2)(x-3) = 0, then the Rational function will be Undefined, that is, the case of Division by Zero (0). So, in the Rational Function f(x) = (x-2)/[(x-2)(x-3)], we see that at x=2 or x=3, the Denominator is equal to Zero (0). But at x=3, we notice that the Numerator is equal to ( 1 ), that is, f(3) = 1/0, hence a Vertical Asymptote at x = 3. But at x=2, we have f(2) = 0/0, 'meaningless'. There is a Hole in the Graph at x = 2.
The x-interceptin is where the graph crosses the y-axis, which is where y equals 0: 0 = 2x + 3 2x = -3 x = -3/2 So the x intercept is as x = -3/2 (and y = 0).
You can do the equation Y 2x plus 3 on a graph. On this graph the Y would equal 5 and X would equal to 0.
It is the point of origin of the x and y axes of the graph
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
a constant y=x-3 dy/dx=0 a straight line
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.
If D > 0 then the graph intersects the x-axis 2 times.If D = 0 then the the x-axis is tangent to the graph.If D < 0 then the graph doe not intersects the x-axis.