# How do you graph y=(3x^2+10x-8)/(x^2+4) using asymptotes, intercepts, end behavior?

Dec 8, 2016

See calculations below

#### Explanation:

The domain of y is ${D}_{y} = \mathbb{R}$

The denominator is ${x}^{2} + 4 > 0$, $\forall x \in \mathbb{R}$

To find the intercepts, let $x = 0$

$y = - \frac{8}{4} = - 2$

The y-intercept is $\left(0 , - 2\right)$

When $y = 0$

$3 {x}^{2} + 10 x - 8 = 0$

$\left(3 x - 2\right) \left(x + 4\right) = 0$

So, $x = \frac{2}{3}$ and $x = - 4$

The x-intercepts are $\left(\frac{2}{3} , 0\right)$ and $\left(- 4 , 0\right)$

To calculate the limits when $x \to \pm \infty$, we take the terms of highest degree in the numerator and the denominator.

${\lim}_{x \to \pm \infty} y = {\lim}_{x \to \pm \infty} \frac{3 {x}^{2}}{x} ^ 2 = 3$

So, the horizontal asymptote is $y = 3$

When $y = 3$, $\implies$, $3 = \frac{3 {x}^{2} + 10 x - 8}{{x}^{2} + 4}$

$3 {x}^{2} + 12 = 3 {x}^{2} + 10 x - 8$

$10 x = 20$, $\implies$. $x = 2$

The curve cuts the horizontal asymptote at $\left(2 , 3\right)$

To find the maximum and minimum, you have to calculate the derivative.

$y = \frac{u}{v}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{u ' v - u v '}{v} ^ 2$

$u = 3 {x}^{2} + 10 x - 8$, $\implies$. $u ' = 6 x + 10$

$v = {x}^{2} + 4$, $\implies$, $v ' = 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(6 x + 10\right) \left({x}^{2} + 4\right) - \left(2 x\right) \left(3 {x}^{2} + 10 x - 8\right)}{{x}^{2} + 4} ^ 2$

$= \frac{6 {x}^{3} + 24 x + 10 {x}^{2} + 40 - 6 {x}^{3} - 20 {x}^{2} + 16 x}{{x}^{2} + 4} ^ 2$

$= \frac{- 10 {x}^{2} + 40 x + 40}{{x}^{2} + 4} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$- 10 \left({x}^{2} - 4 x - 4\right) = 0$

$\Delta = {b}^{2} - 4 a c = 16 + 16 = 32$

$x = \frac{4 \pm \sqrt{32}}{2} = \frac{4 \pm 4 \sqrt{2}}{2} = 2 \pm 2 \sqrt{2}$

So we have 2 points, $x = 2 + 2 \sqrt{2}$ and $x = 2 - 2 \sqrt{2}$

We do a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$2 - 2 \sqrt{2}$$\textcolor{w h i t e}{a a a a}$$2 + 2 \sqrt{2}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$\frac{\mathrm{dy}}{\mathrm{dx}}$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a a a a a a a}$$\downarrow$$\textcolor{w h i t e}{a a a a a a a a a}$$\uparrow$$\textcolor{w h i t e}{a a a a a a a}$$\downarrow$

$x = 2 - 2 \sqrt{2}$ corresponds to a minimum

$x = 2 + 2 \sqrt{2}$ corresponds to a maximum

graph{(y-(3x^2+10x-8)/(x^2+4))(y-3)=0 [-10, 10, -5, 5]}

Dec 8, 2016

x-intercepts $\left(y = 0\right) : - 4 \mathmr{and} \frac{2}{3}$; y-intercept $\left(x = 0\right) : - 2$. Horizontal asymptote: $\leftarrow y = 3 \rightarrow . y \in \left(- 3.04 , 4.04\right)$, nearly.

#### Explanation:

x-intercepts $\left(y = 0\right) : - 4 \mathmr{and} \frac{2}{3}$; y-intercept $\left(x = 0\right) : - 2$

By actual division,

$y = 3 + \frac{10 \left(x - 2\right)}{{x}^{2} + 4} \to \left(2 , 3\right)$ is a point on the graph.

$y = 3 + \frac{10 \left(2 - \frac{1}{x}\right)}{x \left(1 + \frac{4}{x} ^ 2\right)} \to 3$, as x to +-oo#.

It is evident that y = 3 is the horizontal asymptote that cuts the graph

at (2, 3)..

$y ' = \frac{- 10 {x}^{2} + 40 x + 40}{{x}^{2} + 1} ^ 2 = 0$, when x = 2(1+-sqrt2)=-.83 and

4.83,

nearly. These turning points are (-.93, -3.04) and (4.83, 4.04), nearly.

The graph justifies mini/max y at these points as -3.04/4.04,

nearly. y is bounded between these extremes.

graph{y(x^2+4)-3x^2-10x+8=0 [-16.29, 16.24, -8.13, 8.14]}