Reciprocal Quadratic Function Question

[Student123]

This question asks me to graph y = 5/(x²+6x+8) - 4. If that function is 1/f(x), I want to find the equation of f(x). How do I do that?

So far, I got f(x) = 1/(1/5)(x+2)(x+4) - 4. Maybe the image will help.



But I don't know how to incorporate the 4 into the the f(x) equation at the bottom. When I graph the f(x), it doesn't correlate with the original function since the invariant points are not located where they should be (y=-1,+1). Is it possible to get the f(x) equation, or should I do another method to graph the original function?

 

[MarkFL]

Hello, and welcome to FMH!

We are given:

[MATH]y=\frac{5}{x^2+6x+8}-4[/MATH]
Before we can invert this, we should combine terms:

[MATH]y=\frac{5-4(x^2+6x+8)}{x^2+6x+8}=-\frac{4x^2+24x+27}{x^2+6x+8}[/MATH]
Can you proceed now?

 

[Student123]

Thanks, I can proceed now.

 

[Deleted member 4993]

This question asks me to graph y = 5/(x²+6x+8) - 4. If that function is 1/f(x), I want to find the equation of f(x). How do I do that?

So far, I got f(x) = 1/(1/5)(x+2)(x+4) - 4. Maybe the image will help.

View attachment 17860

But I don't know how to incorporate the 4 into the the f(x) equation at the bottom. When I graph the f(x), it doesn't correlate with the original function since the invariant points are not located where they should be (y=-1,+1). Is it possible to get the f(x) equation, or should I do another method to graph the original function?

Did you mean functional inverse [f^-1(x)]or multiplicative inverse [1/f(x)]?

What did the original problem say - EXACTLY?

 

[Student123]

It just told me to graph y = 5/(x²+6x+8) - 4. I don't know what multiplicative inverse is, so I'm guessing it's a functional inverse of a quadratic. I think I figured out the issue though: it shifts down 4, so that the invariant points are instead located on y = +/-1 - 4, and it also affects the vertex. I just got confused on the vertical shift transformation because it shouldn't change the quadratic function that its based on, which changes where the vertical asymptotes are as well. But, thanks anyways.