# How to write reciprocal squared function shifted right by $3$ and down by $4$

Given a reciprocal squared function that is shifted right by $$3$$ and down by $$4$$, write this as a rational function.

I am uncertain how to denote this. My attempt: $$\frac{1}{x^2-3}-4$$

But I need to show this as a rational function. Right now the $$-4$$ is disconnected from the fraction part.

Is it just this?: $$\frac{1}{x^2-3-4}$$ $$\frac{1}{x^2-7}$$

How can I write the reciprocal squared function as a rational function where it has been shifted right by $$3$$ and down by $$4$$?

edit: Is it?: $$\frac{1}{(x-3)^2}+4$$

• $\frac1{x-3}-4\ne\frac1{x-3-4}$ Aug 29, 2020 at 15:12
• What's a reciprocal square function? I think the confusion here stems from the fact that the wording is vague. I really can't guess what is intended. None of your functions reflect the "squared" so I assume they are all wrong, but who knows?
– lulu
Aug 29, 2020 at 15:14
• Is the reciprocal squared function referring to $\frac1{x^2}$? If so, then all your expressions are wrong. Aug 29, 2020 at 15:15
• Also if you want to shift a function $f(x)$ by $b$ units to the right, do $f(x+b)$. If you want to shift a function $g(x)$ by $b$ units down, then do $g(x)-b$. This should be enough information to determine the answer, no matter what your function is. Aug 29, 2020 at 15:19
• I suspect what they mean is the function $f(x) = \frac{1}{(x - 3)^2} - 4$. However, the way the question is phrased makes the sequence of transformations unclear. Note that replacing $x$ by $x - 3$ shifts the graph to the right three units and subtracting $4$ from the expression shifts it down by $4$ units. Aug 29, 2020 at 17:11

$$f$$ is a reciprocal squared function: $$f(x) = \frac{1}{x^2}$$
$$g$$ is $$f$$ shifted by $$a$$ units to the right: $$g(x)=f(x-a)\\g(x)=\frac{1}{(x-a)^2}$$ $$h$$ is $$g$$ shifted by $$b$$ units down $$h(x) = g(x)-b\\h(x)=\frac{1}{(x-a)^2}-b$$ So if you shift $$f$$ by 3 units to the right and 4 units down you would get the following function $$h$$: $$h(x)=\frac{1}{(x-3)^2}-4$$ Now to simplify the expression of $$h$$ or to make it a "rational function" you just have to find the common denominator of the 2 summands which is in this case $$(x-3)^2$$: $$h(x)=\frac{1}{(x-3)^2}-\frac{4(x-3)^2}{(x-3)^2}=\frac{1-4(x^2-6x+9)}{(x-3)^2}\\h(x)=\frac{-4x^2+24x-35}{(x-3)^2}$$
For simplicity call $$u=(x-3)^2$$ so that $$h(x)=1/u + 4 = 1/u + 4u/u=(1+4u)/u$$ and now substituting back in we have $$h(x)=(1+4(x-3)^2)/(x-3)^2$$ which is the quotient of two polynomials as desired.
$$f(x\pm k)$$ shifts a function to the left/right by $$k$$. $$f(x) \pm m$$ shifts a function up/down by $$m$$.
So $$f(x-3) + 4$$ will shift a function to the right by $$3$$ and up by $$4$$.
So if $$f([\color{blue}x]) = \frac 1{[\color{blue}x]^2}$$
then $$f([\color{red}{x-3}])+ 4 = \frac 1{[\color{red}{x-3}]^2} + 4$$