Q: How do you graph the equation y -x 3?

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y=x^3

y = x + 3The graph of that equation is a straight line with a slope of 1 (45Â° rising to the right)and passing through the point y=3 on the y-axis.

y = x - 3The graph is a straight line, passing through the point [ y = -3 ] on the y-axis,and having a slope of [ 1 ].

The is a straight line parallel to the y-axis with an x intercept at -3.

y = -x2 - 4x - 3. The constant, 3, tells us where the graph would hit the y-axis. In this case, it would hit the y-axis at -3. Solving the equation y = -(x2 + 4x + 3) ; y=-(x+3)(x+1) Therefore hits the x-axis at -3 and -1. Since it's a parabola (x2), the graph can either start from the top, or from the bottom. An equation starting with ax2 will start from the top, and an equation with -ax2 will start from the bottom. So therefore, the graph starts from the bottom of the page, goes from -3 and -1 on the x-axis and intercepts the y-axis at -3.

Related questions

y=x^3

To graph the equation y-x=3, first rearrange it in slope-intercept form by isolating y: y=x+3. This equation represents a line with a slope of 1 and y-intercept of 3. You can plot the y-intercept at (0,3) then use the slope to find another point and draw a straight line connecting the two points.

y = x + 3The graph of that equation is a straight line with a slope of 1 (45Â° rising to the right)and passing through the point y=3 on the y-axis.

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.

y = x - 3The graph is a straight line, passing through the point [ y = -3 ] on the y-axis,and having a slope of [ 1 ].

Assuming the graph is: x = - 305 y, then the y intercept is at y = 0Assuming the graph is: x - 3 = 5y, then the y intercept is at y = -3/5The main point for you to realize here is that a graph represents an equation,and " -305y " is not an equation. So some kind of assumption has to be madein order to come up with something that can be graphed.

The given equation y=3+x is a linear line with the slope of 1 and y-intercept of 3.

The is a straight line parallel to the y-axis with an x intercept at -3.

Put a dot on the x axis at x=3 and a dot on the y axis at y=3 and draw a straight line between them.

y = -x2 - 4x - 3. The constant, 3, tells us where the graph would hit the y-axis. In this case, it would hit the y-axis at -3. Solving the equation y = -(x2 + 4x + 3) ; y=-(x+3)(x+1) Therefore hits the x-axis at -3 and -1. Since it's a parabola (x2), the graph can either start from the top, or from the bottom. An equation starting with ax2 will start from the top, and an equation with -ax2 will start from the bottom. So therefore, the graph starts from the bottom of the page, goes from -3 and -1 on the x-axis and intercepts the y-axis at -3.

Yes, for example if you have y=x but you shifted the equation up 3 units hence: y=x+3. than you will receive a different y from every instance (point) of x. Reference: collegemathhelper.com/2015/11/horizontal-graph-transformations-for.html

You could not graph (y-x-2) because it has no equal sign in it. In order to graph an equation, there must be a value that the numbers and variables are equal to. (e.g. y=2x+3)