If the x^4 term is positive, then the curve rises in the 1st and 2nd quadrants, so there has to be at least one minimum point.
if the x4 term is negative, then the curve drops down in the 3rd and 4th quadrants, and there has to be at least one maximum.
For yours the curve will be downwards, so it could look like an upside down W
Draw a WW. You will see that there are 2 mins and 1 max if it opens up, and if the curve looks like an M, (-x^4), there could be 2 max's and 1 min
There cannot be 3 max's and 1 min for a quartic
Your statement about the turning points is correct, but those turning points could be maximums or minimums
I will give you a few examples of equations where it is obvious, I will leave the equations in factored form
1. y = (x+1)(x+2)(x-3)(x+4)
The curve opens up , and has x-intercepts at
x = -1,-2,3, and 4 , so it is easy to see that there must be 2 mins and 1 max
2. y = (x-2)(x-3)(x^2 + 4)
http://www.wolframalpha.com/input/?i=+%28x-2%29%28x-3%29%28x%5E2+%2B+4%29%3D0
3. y = (x-10)(x^3-3)
http://www.wolframalpha.com/input/?i=+%28x-10%29%28x%5E3-3%29%3D0
Play around with different quartics on this webpage, and you can see the different cases.
try:
y = x^4
y = x^4 + x^3
y = x^4 + x^3 + x^2
etc
now change some of the signs .
Have some fun with math.
Determine the maximum and minimum number of turning points for the function h(x) = -2x^4 - 8x^3 + 5x -6.
Maximum:3
Minimum:1
Is this a valid reason:
A quartic polynomial function has a 3 Turning points. The turning point is always 1 less than the degree. For example degree 4=3 TP, degree 5=4 turning points?
What about the minimum?
1 answer
1 answer