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

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    $\begingroup$ $\frac1{x-3}-4\ne\frac1{x-3-4}$ $\endgroup$
    – Aiden Chow
    Aug 29, 2020 at 15:12
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    $\begingroup$ 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? $\endgroup$
    – lulu
    Aug 29, 2020 at 15:14
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    $\begingroup$ Is the reciprocal squared function referring to $\frac1{x^2}$? If so, then all your expressions are wrong. $\endgroup$
    – Aiden Chow
    Aug 29, 2020 at 15:15
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    $\begingroup$ 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. $\endgroup$
    – Aiden Chow
    Aug 29, 2020 at 15:19
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    $\begingroup$ 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. $\endgroup$ Aug 29, 2020 at 17:11

3 Answers 3

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$f$ 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 $(x-3)^2$: $$h(x)=\frac{1}{(x-3)^2}-\frac{4(x-3)^2}{(x-3)^2}=\frac{1-4(x^2-6x+9)}{(x-3)^2}\\h(x)=\frac{-4x^2+24x-35}{(x-3)^2}$$

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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.

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$f(x\pm k)$ 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 $f([\color{red}{x-3}])+ 4 = \frac 1{[\color{red}{x-3}]^2} + 4$

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