# How do you graph f (x) = (24x – 156) /( 36 – x^2)?

Jun 5, 2015

First of all, we have to define the qualitative characteristics of this function. Obviously, it has asymptotes at points where the denominator equals to zero, that is where $36 - {x}^{2} = 0$ or ${x}^{2} = 36$ or $x = \pm 6$.

Here is one approach to graph this function (and there are others).

$f \left(x\right) = \frac{24 x - 156}{36 - {x}^{2}} = \frac{24 \left(x - 6\right) - 12}{36 - {x}^{2}} =$

$= \frac{24 \left(x - 6\right)}{\left(6 - x\right) \left(6 + x\right)} - \frac{12}{36 - {x}^{2}} =$

$= - \frac{24}{x + 6} - \frac{12}{36 - {x}^{2}}$

Now we have to construct two separate graphs:
${f}_{1} \left(x\right) = - \frac{24}{x + 6}$ and ${f}_{2} \left(x\right) = - \frac{12}{36 - {x}^{2}}$
The graph of the second function can be constructed by inverting the graph of a function $g \left(x\right) = 36 - {x}^{2}$, which is a parabola with "horn" directed down and intersecting the X-axis at points ${x}_{1} = 6$ and ${x}_{2} = - 6$. So, wherever $g \left(x\right)$ goes to infinity, ${f}_{2} \left(x\right)$ goes to zero and vice versa.
Here is a graph of $g(x)=36-x^2: graph{36-x^2 [-80, 80, -40, 40]} And inverted graph of f_2(x)=-12/(36-x^2) graph{-12/(36-x^2) [-80, 80, -40, 40]} Adding f_1(x)$\mathmr{and}$f_2(x), we see that (1) around point x=-6, where both functions go to positive infinity if approaching this point from the left and to negative infinity if approaching from the right, the asymptotic behavior would be exactly as that of each component (-> +oo$\mathfrak{o} m t h e \le f t \mathmr{and}$->-oo from the right) (2) around point x=6$f u n c t i o n$f_1(x)$i s \lim i t e d w h i \le$f_2(x)$h a s v e r t i c a l a s y m p \to t e , s o t h e \sum o f t h e s e f u n c t i o n w o \underline{d} b e h a v e l i k e$f_2(x)$, t \hat{i} s$-> -oo$\mathfrak{o} m t h e \le f t \mathmr{and}$->+oo from the right. (3) in between point x=-6$\mathmr{and}$x=6$f u n c t i o n$f_1(x)$i s f \in i t e \mathmr{and} f u n c t i o n$f_2(x) just adds a small correction to it. So, they sum would be finite (staying negative, as both components are negative) (4) to the left of x=-6$\bot h c o m p o \ne n t s b e h a v e s i m i l a r l y , s o t h e i r \sum w o \underline{d} r e s e m b \le e a c h o \ne o f t h e m , t \hat{i} s$->0$a s$x->-oo. (5) to the right of x=6 f_2(x)$b e c o m e s \neg l i \ge n t l y s m a l l a s$x->+oo$, s o t h e \sum o f t h e s e t w o c o m p o \ne n t s w o \underline{d} r e s e m b \le t h e f i r s t o \ne ,$f_1(x)#. The resulting graph would look like this: graph{-24/(x+6)-12/(36-x^2) [-50, 50, -40, 40]} Jun 5, 2015 First of all, we have to define the qualitative characteristics of this function. Obviously, it has asymptotes at points where the denominator equals to zero, that is where $36 - {x}^{2} = 0$or ${x}^{2} = 36$or $x = \pm 6$. Here is one approach to graph this function (and there are others). $f \left(x\right) = \frac{24 x - 156}{36 - {x}^{2}} = \frac{24 \left(x - 6\right) - 12}{36 - {x}^{2}} =$$= \frac{24 \left(x - 6\right)}{\left(6 - x\right) \left(6 + x\right)} - \frac{12}{36 - {x}^{2}} =$$= - \frac{24}{x + 6} - \frac{12}{36 - {x}^{2}}$Now we have to construct two separate graphs: ${f}_{1} \left(x\right) = - \frac{24}{x + 6}$and ${f}_{2} \left(x\right) = - \frac{12}{36 - {x}^{2}}$and add them up. Graph of the first function is a hyperbola shifted by 6 units to the left, stretched by a factor of 24 and inverted the sign by reflecting it relative to the X-axis. graph{-24/(x+6) [-40, 40, -20, 20]} The graph of the second function can be constructed by inverting the graph of a function $g \left(x\right) = 36 - {x}^{2}$, which is a parabola with "horn" directed down and intersecting the X-axis at points ${x}_{1} = 6$and ${x}_{2} = - 6$. So, wherever $g \left(x\right)$goes to infinity, ${f}_{2} \left(x\right)$goes to zero and vice versa. Here is a graph of $g \left(x\right) = 36 - {x}^{2}$: graph{36-x^2 [-80, 80, -40, 40]} And inverted graph of ${f}_{2} \left(x\right) = - \frac{12}{36 - {x}^{2}}$graph{-12/(36-x^2) [-80, 80, -40, 40]} Adding ${f}_{1} \left(x\right)$and ${f}_{2} \left(x\right)$, we see that (1) around point $x = - 6$, where both functions go to positive infinity if approaching this point from the left and to negative infinity if approaching from the right, the asymptotic behavior would be exactly as that of each component ($\to + \infty$from the left and $\to - \infty$from the right) (2) around point $x = 6$function ${f}_{1} \left(x\right)$is limited while ${f}_{2} \left(x\right)$has vertical asymptote, so the sum of these function would behave like ${f}_{2} \left(x\right)$, that is $\to - \infty$from the left and $\to + \infty$from the right. (3) in between point $x = - 6$and $x = 6$function ${f}_{1} \left(x\right)$is finite and function ${f}_{2} \left(x\right)$just adds a small correction to it. So, they sum would be finite (staying negative, as both components are negative) (4) to the left of $x = - 6$both components behave similarly, so their sum would resemble each one of them, that is $\to 0$as $x \to - \infty$. (5) to the right of $x = 6$${f}_{2} \left(x\right)$becomes negligently small as $x \to + \infty$, so the sum of these two components would resemble the first one, ${f}_{1} \left(x\right)\$.