P(x) = -(x-a)^4 (x-b)^3 (x-c)
where a>0, b<0, c>0, a<c
How do I sketch a graph of this?
Thanks
First you put the points a, b and c on the x-axis, doesn't matter precisely where, as long as it satisfies the conditions: a>0, b<0, c>0, a<c.
Then you use the fact that at a, b and c the function is zero. At b and c, the function changes sign, so it crosses the x-axis, but at x = a, there is no change of sign, there the function just touches the x-axis.
At x = c, the function crosses the x-axis at a nonzero angle. But at x = b, the fuction crosses the x-axis at zero angle. I.e. the function touches the x-axis there but unlike at point x = a, it does cross the x-axis.
To sketch the graph, you can follow these steps:
1. Plot the points a, b, and c on the x-axis. Since a>0, b<0, and c>0, make sure to place them accordingly.
2. Determine the behavior of the function near each point. At point a, the function touches the x-axis without crossing it, so it will be a local minimum.
3. At points b and c, the function changes sign and crosses the x-axis. Since the powers of (x-b) and (x-c) are odd (3 and 1, respectively), the function will cross the x-axis at these points.
4. Analyze the end behavior of the function. Since the powers of (x-a), (x-b), and (x-c) are all even, the graph will approach the x-axis on both sides as x approaches positive or negative infinity.
5. Determine the multiplicity of each factor. The factor (x-a) has a multiplicity of 4, (x-b) has a multiplicity of 3, and (x-c) has a multiplicity of 1. This information will help in understanding the shape of the graph near these points.
6. Sketch the graph by considering these factors. The shape of the graph near point a will be a smooth curve that touches the x-axis and turns upwards. Near points b and c, the graph will cross the x-axis at different angles.
Remember to label the x and y-axis and any other important points on the graph.
Note: This explanation assumes you are familiar with basic graphing concepts and terminology.