For the function f(x)=3(x^2-8)(x^2-9)^2:

Question

 a)List each real zero and its multiplicity. b) Determine whether the graph crosses or touches the x-axis at each x-intercept c) Determine the max number of turning points d) Determine the end behavior.

Answer 1

If we factor this out completely,

 

f(x) = 3(x² - 8)(x - 3)(x + 3)(x - 3)(x + 3)

 

 

The factor      x² - 8 = 0

 

                     x² =8

                    x = ±2√2

 

has 2 zeros        x = -2√2    and      x = 2√2.  They both cross the x-axis.

 

 

The factors   (x - 3)     and     (x + 3)   have the zeros

 

x = -3   with a multiplicity of 2

x = 3    with a multiplicity of 2

 

 

Overall, the graph crosses the x-axis twice and touches the x-axis twice.

 

To find the max number of turning points, we list the zeros in order and name their multiplicities,

 

x = -3       ,   x = -2√2    ,      x = 2√2    ,      x = 3

 

 

When the graph goes up to hit x=-3 it starts to decrease. One turning point.  Then goes back up to hit x=-2√2.  Back down to hit x=2√2.   Up to hit x=3 then back down.   That is 5 turning points max.

 

 

We have a polynomial of degree 6. 

 

 

As for the end behavior, the function has a positive leading coefficient with odd degree.  As x decreases, f(x) increases.  As x increases, f(x) increases.