Posted by Jen on .
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?
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.