How do you graph $\frac{4 x + 1}{x^{2} - 1}$ $(4x+1)/(x^2-1$ using asymptotes, intercepts, end behavior?

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Answer 1 · 2017-11-09T16:17:09

See below.

$\frac{4 x + 1}{x^{2} - 1}$ $(4x+1)/(x^2-1)$ is undefined for $x = 1 or - 1$ $x=1 or -1$ ( Division by zero )

So the line $x = 1 and x = - 1$ $x=1 and x=-1$ are vertical asymptotes.

Y axis intercept occurs when $x = 0$ $x =0$

$y = \frac{4 (0) + 1}{{(0)}^{2} - 1} = - 1$ $y=(4(0)+1)/((0)^2-1)=-1$ ( 0 , -1 )

x axis intercepts occur when y = 0.

$\frac{4 x + 1}{x^{2} - 1} = 0$ $(4x+1)/(x^2-1)=0$

We can solve this by reasoning that if $(4 x + 1) = 0$ $(4x+1)=0$

Then:

$\frac{4 x + 1}{x^{2} - 1} = 0$ $(4x+1)/(x^2-1)=0$

$:.$

$(4x+1)=0 =>x=-1/4$

Note: Do not try to solve for $(x^2-1)=0$ ( zero denominator is invalid )

x axis intercept at:

$(-1/4 , 0 )$

For limits to infinity we only need to concern ourselves with the two terms containing the variable.

$(4x)/x^2$

$(4x)/x^2=4/x$

$lim_(x->oo)(4/x)=0$

$lim_(x->-oo)(4/x)=0$

So the x axis is a horizontal asymptote.

Graph:

enter image source here

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