# How do you find the domain and range of f(x)=(12x)/(x^2-36)?

Jul 16, 2018

Below

#### Explanation:

Looking at the graph, you can immediately see that there are 2 vertical asymptotes because ${x}^{2} - 36 \ne 0$ so $x = \pm 6$ are the vertical asymptotes. Therefore, the graph cannot have the points with the x-coordinates $x = 6$ and $x = - 6$

The horizontal asymptote is $y = 0$ since the degree of the numerator is less than the degree of the denominator. This is because if you imagine letting $x$ be any number, then ${x}^{2} - 36$ will be a whole lot bigger than $12 x$ and since the small number divided by a larger number, then $\frac{12 x}{{x}^{2} - 36} \to 0$

Therefore, the graph cannot have the points with the y-coordinate $y = 0$

However, what asymptotes really tell you about the graph is that the end points of the graph will be approaching the horizontal and vertical asymptotes but they will never touch the asymptotes. Basically, it tells about the shape of the graph which can help you determine the domain and range of the graph.

Intercepts
When $y = 0$, $x = 0$
When $x = 0$, $y = 0$
You will notice that the graph can pass through $\left(0 , 0\right)$ but the endpoints of the graph will be approaching $y = 0$ and not cross $y = 0$. This is because asymptotes affect the end points only.

Hence,
Domain: all reals $x$ except when $x = \pm 6$
Range: all reals $y$

Below is the graph

graph{(12x)/(x^2-36) [-10, 10, -5, 5]}

Jul 16, 2018

The domain is x in (-oo, -6)uu-6,6)uu(6,+oo). The range is $y \in \mathbb{R}$

#### Explanation:

The denominator must be $\ne 0$

Therefore,

${x}^{2} - 36 \ne 0$

$\left(x + 6\right) \left(x - 6\right) \ne 0$

$x \ne - 6$ and $x \ne 6$

The domain is x in (-oo, -6)uu-6,6)uu(6,+oo)

To find the range, let

$y = \frac{12 x}{{x}^{2} - 36}$

$y \left({x}^{2} - 36\right) = 12 x$

$y {x}^{2} - 12 x - 36 y = 0$

This is a quadratic equation in $x$ and in order to have solutions, the discriminant $\ge 0$

Therefore,

$\Delta = {\left(- 12\right)}^{2} - 4 \left(y\right) \left(- 36 y\right)$

$= 144 + 144 {y}^{2} \ge 0$

$\implies$, $144 \left(1 + {y}^{2}\right) \ge 0$

Therefore,

$\forall y \in \mathbb{R} , \Delta > 0$

The range is $y \in \mathbb{R}$

graph{12x/(x^2-36) [-32.49, 32.46, -16.24, 16.25]}