Posted by Em on Tuesday, October 23, 2007 at 10:55pm.
Determine the maximum possible number of turning points for the graph of the fucntion.
g(x)=1/5x + 2
I got 0
How do I graph f(x)=x^54x^312x

Functions  Reiny, Tuesday, October 23, 2007 at 11:09pm
your first one is right
for f(x)=x^54x^312x
=x(x^4  4x^2  12)
=x(x^26)(x^2+2)
if this had been the equation
x^54x^312x = 0
there would be solutions of
x=0, x=±√6 = ±2.45, and 2 complex roots
so the graph crosses at 2.45, 0, 2,45
it rises into the first quadrant, and drops into the third quadrants.
The two complex roots cause a "wiggle" in the first quadrant above the xaxis