y = x2 is an (approximately) U shaped graph that is entirely above the x axis and is symmetric about the y axis. y = x3 is asymptotically negatively infinite when x is negatively infinite and positively infinite when x is positively infinite. It is symmetric about the line x+y=0.
This depends on what you mean by "X3". If the equation is x3-x, then the graph would increase up to about y=0.385, crossing the x-axis at x=-1, then decrease to y=-0.385, crossing the origin, then begin to increase again, crossing the x-axis at x=1. If you just mean x*3-x, then the answer would be 2x. You just enter the function in as f(x)=x^3-x and it will tell you the graph. For future reference, x3 is x^3, not x3.
x3
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
Just label one axis x and the other axis y. Voila!! x y graph!
Graph the following Inequalities: x > 3
y = x2 is an (approximately) U shaped graph that is entirely above the x axis and is symmetric about the y axis. y = x3 is asymptotically negatively infinite when x is negatively infinite and positively infinite when x is positively infinite. It is symmetric about the line x+y=0.
This depends on what you mean by "X3". If the equation is x3-x, then the graph would increase up to about y=0.385, crossing the x-axis at x=-1, then decrease to y=-0.385, crossing the origin, then begin to increase again, crossing the x-axis at x=1. If you just mean x*3-x, then the answer would be 2x. You just enter the function in as f(x)=x^3-x and it will tell you the graph. For future reference, x3 is x^3, not x3.
x3
A line graph needs an equation. x-2 and x3 are expressions: neither is an equation.
To translate the graph y = x to the graph of y = x - 6, shift the graph of y = x down 6 units.
The graph is a diagonal line with a slope of 1, passing through the x-axis at (3, 0) and the y-axis at (0, -3), extending from the lower left to the upper right.
x6 - y6 = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x + y)(x2 - xy + y2)(x - y)(x2 + xy + y2)
what is the anwer for x=3 and x=2
Yes, y = -x3+1 is a function. You can graph it and see that it passes the vertical line test. See related link, below.
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.
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)