|
Question 1054657: Write a polynomial function that has the following characteristics?
Extends from quadrant II to IV
Y-intercept of 6
One x-intercept is at 2
2 turning points
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! Since the function must have two turning points, it must be at least of degree 3.
It crosses the x-axis at x=2, which means one of the factors must be (x-2)
Also, the constant term, when x=0, is equal to 6.
From this information we can write:
f(x) = (x-2)(ax^2+bx-3)
The function starts out in quadrant II and ends in quadrant IV,
so for negative values of x, it must be >0 and for positive values of x, it must be <0.
This implies the coefficient on the leading term is negative.
Let us assume that the quadratic term of f(x) = 0
Thus we need to find a and b such that x-2 is a factor of ax^3+bx+6
If we perform the division, we are left with the requirement that b = -4a - 3
At the turning points, the df/dx = 0, and there must be two.
df/dx = 3ax^2+b = 0 -> x = +- sqrt(-b/3a) = +- sqrt((4a+3)/3a)) = +- sqrt(4/3+1/a), which must be >0 for there to be two real roots.
Therefore a must be less than -3/4.
Let a = -1, then b = -4(-1) - 3 = 1
So a function that satisfies the requirements is
f(x) = -x^3 + x + 6
|
|
|
| |
