# How to get a whole number from $y = \frac{1}{x + 2}$ How to come up with a whole number for y.

I keep coming up with fractions from $y = \frac{1}{x +2}$

I've tried numerous numbers, as in, $1, 2, 3 ,4 , 5, -1, -2, -3, -4, -5$.

For example, $y = \frac{1}{1 + 2} = \frac{1}{3}$.

It's suppose to make two separate identical curves on a Cartesian plane, and one is positive and negative, the other is in the negatives. When the $x$ values are positive, the $y$ value is less then $1$. When the $x$ value is negative the $y$ value goes greater then $1$.

• Is $x$ limited to be an integer? – peterwhy Jul 9 '16 at 21:15
• Try letting $x=\frac{1}{n}-2$ where $n$ is any whole number (other than 0). – user84413 Jul 9 '16 at 21:15
• Your edited statement is very ... mystifying. What is "it", what curves are you referring to, what graph are you referring to? – pjs36 Jul 9 '16 at 21:18
• @peterwhy yes $x$ is limited to an integer. – William Zlacki Jul 9 '16 at 21:19

Because the numerator is $1$ in this case, using whole numbers for $x$ the only way that the fraction $\frac{1}{x+2}$ can be a whole number is when the denominator is either $1$ or $-1$. Knowing this, the only solutions are when $x$ is either $-1$ or $-3$. This gives us $y$ values of $1$ and $-1$ respectively.
If you just want $y$ to be an integer, but you don't need $x$ to be an integer, you could first solve your function for $x$ like this: $$y = \frac{1}{x+2} \qquad\implies\qquad x = \frac{1-2y}{y}$$ Now supposing you want the $y$-value of your original function to be $y=5$, you just evaluate this new form at $y=5$ and find $x$, so $$x = \frac{1-2y}{y} = \frac{1-2(5)}{(5)} = -\frac{9}{5}\;.$$ So the $x$-value of $-9/5$ evaluates to $5$ when used in your original function.
• What exactly do you mean that the "graph goes past $1$?" Sure the graph keeps going, but none of the $y$-values of the graph for $x>-1$ will be whole numbers: they will just be some real numbers between $1$ and $0$. The only points on the graph where both the $x$-value and $y$-value are integers are at $(-1,1)$ and $(-3,-1)$. – Mike Pierce Jul 9 '16 at 21:16