Finding Equation of Polynomial Function

[wind_surfer]

Hello, I've been stuck on this question for a while, the question asks me to graph y=1/f(x) and I am given y=f(x) in graph form (see picture below). I have been trying to find the equation of this graph by using the x-intercepts and finding the "a" in f(x)=a(x+4)(x)(x-6) though I can't seem to find an "a" that allows my equation to match the graph I'm given. The most accurate equation I have come up with is f(x)=-0.125(x+4)(x)(x-6) however it does not go through the point (-2,-2) like it does in the original graph. What am I doing wrong?Is there something I am missing? Any help would be appreciated.


Thankyou,

 

[wjm11]

f(x)=a(x+4)(x)(x-6)

Plug in (-2,-2):
-2 = a(-2+4)(-2)(-2-6)
-2 = a(2)(-2)(-8)
-2 = a(32)
a = -2/32 = -1/16
(You lost a factor of 2 somewhere.)

 

[wind_surfer]

But I still get the same problem.
-2 = a(-2+4)(-2)(-2-6)
-2 = a(2)(-2)(-8)
-2 = a(32)
a = -2/32 = -1/16
When I plug in (4,8)
8 = a(4+4)(4)(4-6)
8 = a(8)(4)(-2)
8 = a(-64)
a = 8/-64 = -1/8
So did I do something wrong when I plugged in (4,8)? I'm still confused..

 

[wjm11]

...the question asks me to graph y=1/f(x)...

Your approach for finding "a" is fine, but perhaps you don't need to know "a" exactly. Perhaps the assumption that both (-2,-2) and (4,8) are points on the curve is incorrect -- thus the discrepancy in calculated "a" values. Is the question just asking for a "sketch" of 1/f(x)? You can do that without knowing "a" exactly. What will happen to 1/f(x) at x = 0, 6, and -4? Will the sign of f(x) and 1/f(x) be the same or different at other values of x?

You should be able to sketch the new function after thinking about those questions.

(You can also plot both graphs in a graphing calculator if you need more help in seeing what is going on.)

Hope that helps.

 

[wind_surfer]

Thankyou for your reply, I think you are right. I was probably looking into the details too much when all they wanted was a sketch and the asymptotes.