# How do you sketch the curve y=(x+1)/sqrt(5x^2+35) by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

Dec 13, 2017

see below
graph{(x+1)/(sqrt(5x^2+35)) [-10, 10, -5, 5]}

#### Explanation:

$f \left(x\right) = \frac{x + 1}{\sqrt{5 {x}^{2} + 35}}$

Vertical asymptote is in the point where it isn't defined that is:
$\sqrt{5 {x}^{2} + 35} \ne 0$
$5 {x}^{2} + 35 \ne 0$
${x}^{2} \ne - \frac{35}{5}$
$x \ne \sqrt{- \frac{35}{5}}$
As you can see this is absolute nonsense. That can't be done. Square root must be $> 0$. That means it doesn't have a vertical asymptote and The Domain is $\mathbb{R}$

In order to determine maximum and minimum we need first derivative:
$f \left(x\right) = \frac{x + 1}{\sqrt{5 {x}^{2} + 35}}$

${f}^{'} \left(x\right) = \frac{{\left(x + 1\right)}^{'} \left(\sqrt{5 {x}^{2} + 35}\right) - \left(x + 1\right) {\left({\left(5 {x}^{2} + 35\right)}^{\frac{1}{2}}\right)}^{'}}{\sqrt{5 {x}^{2} + 35}} ^ 2$

${f}^{'} \left(x\right) = \frac{1 \cdot \sqrt{5 {x}^{2} + 35} - \left(x + 1\right) \frac{\left(1 \cdot 10 x\right)}{2 \sqrt{5 {x}^{2} + 35}}}{\sqrt{5 {x}^{2} + 35}} ^ 2$

${f}^{'} \left(x\right) = \frac{\frac{2 {\left(\sqrt{5 {x}^{2} + 35}\right)}^{2} - 10 x \left(x + 1\right)}{2 \sqrt{5 {x}^{2} + 35}}}{\sqrt{5 {x}^{2} + 35}} ^ 2$

f^'(x)=(2(5x^2+35)-10x(x+1))/(2(sqrt(5x^2+35))^3

f^'(x)=(cancel(10x^2)+2*35cancel(-10x^2)-2*5x)/(2(sqrt(5x^2+35))^3

f^'(x)=(cancel2(35-5x))/(cancel2(sqrt(5x^2+35))^3

${f}^{'} \left(x\right) = \frac{5 \left(7 - x\right)}{\sqrt{5 {x}^{2} + 35}} ^ 3$

$\sqrt{5 {x}^{2} + 35}$ is always positive so we don't care about that. However $\left(7 - x\right)$ can be + and -
$x \in \left(- \infty , 7\right) \Leftrightarrow {f}^{'} \left(x\right) > 0 \implies$ f goes up
$x \in \left(7 , \infty\right) \Leftrightarrow {f}^{'} \left(x\right) < 0 \implies$ f goes down

maximum is in x=7$\quad f \left(7\right) = \frac{7 + 1}{\sqrt{5 \cdot {\left(7\right)}^{2} + 35}} = \frac{8}{\sqrt{280}} = \frac{4}{\sqrt{70}} \approx 0.478$

function doesn't have minimum

horizontal asymptote: $L i {m}_{x \rightarrow \pm \infty} f \left(x\right)$
1. for $+ \infty$
$L i {m}_{x \rightarrow \infty} \frac{\cancel{x} \left(1 + \frac{1}{x}\right)}{\cancel{x} \left(\sqrt{5 + \frac{35}{{x}^{2}}}\right)} = L i {m}_{x \rightarrow \infty} \frac{1 + \frac{1}{x}}{\sqrt{5 + \frac{35}{{x}^{2}}}} = \frac{1 + 0}{\sqrt{5 + 0}} = \frac{1}{\sqrt{5}} \approx 0.447$

1. for $- \infty$
$L i {m}_{x \rightarrow - \infty} f \left(x\right) = \text{by implementing L'Hopitals rule we get} = L i {m}_{x \rightarrow - \infty} \left(\sqrt{5 {x}^{2} + 35} \cdot 5 x\right) = - \infty$

Intercepts:
$\mathmr{if} x = 0 \quad \implies \quad y = \frac{1}{\sqrt{35}}$

$\mathmr{if} y = 0 \quad \implies \quad 0 = x + 1 \quad \implies \quad x = - 1$