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?:$$\frac{1}{(x-3)^2}+4$$
$\endgroup$ 93 Answers
$\begingroup$$f$ is a reciprocal squared function: $$ 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}$$
For simplicity call $u=(x-3)^2$ so that $h(x)=1/u + 4 = 1/u + 4u/u=(1+4u)/u$ and now substituting back in we have $h(x)=(1+4(x-3)^2)/(x-3)^2$ which is the quotient of two polynomials as desired.
$\endgroup$ $\begingroup$$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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