Thread: How to find the x-intercept of this polynomial function?

F(X) = -2(x+1/2)^2(x+4)^3

Where -2(x+1/2)^2 = 0 & (X+4)^3 = 0

Thank you in advanced for any help!

In a larger picture, I am actually trying to figure out whether the graph crosses or touches the x-axis at each x-intercept. Teaching myself out of the book is proving to be a real pain...

EDIT: I want to say that the X intercepts are X=-1/2 and X=-4 but I am not positive. is the -2, ^2, and ^3 insignificant when finding the x intercepts?

Also.. Since the problem as a whole is a degree of 5, does this mean that the function will have 5 turning points?

Originally Posted by duBxStep

F(X) = -2(x+1/2)^2(x+4)^3

Where -2(x+1/2)^2 = 0 & (X+4)^3 = 0

Originally Posted by duBxStep

In a larger picture, I am actually trying to figure out whether the graph crosses or touches the x-axis at each x-intercept. Teaching myself out of the book is proving to be a real pain...

EDIT: I want to say that the X intercepts are X=-1/2 and X=-4 but I am not positive. is the -2, ^2, and ^3 insignificant when finding the x intercepts?

To determine the x-intercept only the value in brackets must be zero: If you multiply zero or take zero to any power that will not change the value.

In addition: A product equals zero if at least one factor equals zero.

Also.. Since the problem as a whole is a degree of 5, does this mean that the function will have 5 turning points?

In general: A polynomial function of degree n has
- a maximum number of zeros of n
- a maximum number of turning points (n-1)
- a maximum number of points of inflection (n-2)

If you have a "double zero" (that means the factor which becomes zero is a square) then the x-intercept is simultaneously a turning point. The graph of the function touches the x-axis.

If you have a "triple zero" (that means the factor which becomes zero is a cube) then the x-intercept is simultaneously a point of inflection with the x-axis as a tangent. The graph of the function crosses the x-axis.