How do you graph f(x)=(2x-1)/(x^3-9x) using holes, vertical and horizontal asymptotes, x and y intercepts?

Nov 14, 2017

x intercept:
$A = \left(\frac{1}{2} , 0\right)$

y intercept:
$N a N$

vertical asymptotes:
$x = 0$
$x = 3$
$x = - 3$

horizontal asymptotes:
$y = 0$ (for both x very big positive and very big negative)

Explanation:

FIRST thing we ALWAYS do is making the function easier:

f_((x))=(2x-1)/((x^3-9x)
$\implies \frac{2 x - 1}{x \left({x}^{2} - 9\right)}$
$\implies \frac{2 x - 1}{x \left({x}^{2} - 9\right)}$
=> (2x-1)/(x(x-3)(x+3) because $\left({a}^{2} - {b}^{2}\right) = \left(a - b\right) \left(a + b\right)$

Now we can "easily" see that $x$ MUST NOT be 0,3,-3 because we know that we must not divided by 0.
We can also see that $\left(2 x - 1 = 0\right)$ if $\left(x = \frac{1}{2}\right)$

vertical asymptotes:

Therefore we now know that the function has 3 vertical asymptotes (we get them when $x = 0$ in the "down side of the equation"):
$x = 0$
$x = 3$
$x = - 3$

horizontal asymptotes:
(I don't know if you know how to use $\lim$ so I will explain in an alternative way)

horizontal asymptotes is a question of a kind of "What does the function "act" when $x$ is REALLY HUGE (and/or negative HUGE)?

To solve it easily, lets check with a calculator what happens when x comes is really big, for example I will check the positives:

${f}_{\left(1 , 000 , 000\right)} = 1.999999 \cdot {10}^{-} 12 = 0.0000000000000001999$
${f}_{\left(1 , 00 , 000 , 000\right)} = 1.9999 \cdot {10}^{-} 15 = 0.0000000000000000001999$

So, the function goes to $y = 0$ for very big x-es.
On the same way, we can chek the very negative x-es:
${f}_{\left(1 , 000 , 000\right)} = 2.000001 \cdot {10}^{-} 12 = 0.00000000000000020001$

So, in that case, the function goes to $y = 0$ for very negative big x-es.

therefore, the horizontal asymptote is:
$y = 0$

x and y intercepts:
Now that we have an "easier" function:
$\left({f}_{\left(x\right)} = \frac{2 x - 1}{x \left(x - 3\right) \left(x + 3\right)}\right)$

We check both "x=0" and y=0":
${f}_{\left(0\right)} = \frac{2 \cdot 0 - 1}{0 \cdot \left(0 - 3\right) \left(0 + 3\right)} = - \frac{1}{0} = N a N$
But we already knew it because $x = 0$ is one of ours' vertical asymptotes...
Now lets check what if ${f}_{\left(x\right)} = 0$ what is x?
$\frac{2 \cdot x - 1}{x \cdot \left(x - 3\right) \left(x + 3\right)} = 0$
$\implies \left(2 \cdot x - 1\right) = 0$ (for x in not 1,3,-3)
$\implies 2 x = 1$
$\implies x = \frac{1}{2}$

So we have only one point $A = \left(\frac{1}{2} , 0\right)$