# How do I graph this?

I am just a high school student, so please excuse me if this question sound silly to you.

How do I graph this : $$\dfrac{9}{x} - \dfrac{4}{y} = 8$$

I could do trial and error, but is there a more systematic way that I can use to graph this equation.

• Can you solve for $y$ ? – dmtri Mar 9 '19 at 11:11
• Please explain what you mean by "graph this equation." Do you mean get it into a form so you can use a graphing calculator or program to show the graph, or do you mean graph it manually without a calculator, or do you mean something else? Could calculus methods such as intervals where the graph increases or decreases be used? How about end behavior? Could/should you use the transformation of a standard graph using translations and expansions? – Rory Daulton Mar 9 '19 at 11:32

I'll assume you want to graph that equation without a graphing calculator or graphing program, and that you want to use only pre-calculus techniques.

So we want to transform the equation into something resembling one of the "standard equations" that we have seen before. First, we simplify the equation by removing the fractions and moving all terms to one side of the equation.

$$\frac 9x - \frac 4y = 8$$ $$\left(\frac 9x - \frac 4y\right)xy = 8xy$$ $$9y - 4x = 8xy$$ $$8xy + 4x - 9y = 0$$

That left side looks factorable. We'll make the first coefficient into one.

$$\frac 18(8xy + 4x - 9y)= \frac 18\cdot 0$$ $$xy + \frac 12x - \frac 98y = 0$$

Now that looks much like factoring a polynomial. We see that we can split the $$\frac 12$$ and the $$-\frac 98$$ by making the left side into the product of two binomials. But first we need to add the appropriate constant term.

$$xy + \frac 12x - \frac 98y + -\frac 98\cdot\frac 12 = -\frac 98\cdot\frac 12$$ $$\left(x-\frac 98\right) \left(y+\frac 12\right) = -\left(\frac 34\right)^2$$

Now we see this is a transformation of the graph of the standard equation

$$xy=1$$

This is a hyperbola. Our graph has a negative number on the right, so it is reflected in an axis, so the branches of the hyperbola are in the upper-left and lower-right areas of the graph. The right side is not one so the vertices of the hyperbola have an x-distance and a y-distance of $$\frac 34$$ from the center. The graph is translated so the vertical asymptote is the line $$x=\frac 98$$ and the horizontal asymptote is the line $$y=-\frac 12$$.

One last thing. We started all this by multiplying by $$xy$$ to remove the fractions. That means that we must pay attention to the points where $$x=0$$ or $$y=0$$. The original equation shows that such points are not on the graph. So even though the origin is included in our final equation, it is not on the original graph. So our hyperbola has a hole at the origin.

That information makes it easy to sketch the graph. We can confirm this by graphing one of the equations on a graphing calculator. Here is $$8xy+4x-9y=0$$ on the TI-Nspire CX scratchpad screen. Except for the hole at the origin, this agrees with my analysis.

I would write $$\frac{9}{x}-8=\frac{4}{y}$$ so, $$\frac{9-8x}{x}=\frac{4}{y}$$ and so, $$y=\frac{4x}{9-8x}$$ and then use a function plotter from the internet.

Yes, there is. Note that\begin{align}\frac9x-\frac4y=8&\iff\frac4y=\frac9x-8\\&\iff y=\frac4{\frac9x-8}=\frac{4x}{9-8x}=-\frac12+\frac9{18-16x}.\end{align}So, graph the function $$x\mapsto-\dfrac12+\dfrac9{18-16x}$$.

Notice the coefficients of $$\frac{1}{x}$$ and $$\frac{1}{y}$$, we can alter $$x$$ in multiples of $$3$$ or $$9$$ to get corresponding values of $$y$$. Also $$x \neq0$$, and $$y\neq 0$$, to satisfy the equation: $$\frac{9}{x}-\frac{4}{y}=8$$

Rewrite the equation as : $$y=\frac{4}{\frac{9}{x}-8}$$ and just put in values of $$x$$ as $$\pm1,\pm2,\pm3,\pm9,\pm12,\pm15,\pm18$$ as well as $$\pm \frac{1}{2},\pm \frac{1}{3}, \pm \frac{1}{9}, \pm \frac{1}{12}$$ to get a rough idea of the curve.

Note there will be a break in the curve at $$x=\frac{9}{8}$$. You can also compute the derivatives and verify in which part is the function increasing or decreasing. To plot it you can use MATLAB or matplotlib in python. Hope this helps....