Mathematics, 15.07.2019 11:00 smariedegray

Graph all vertical and horizontal asymptotes of the function f(x)= -8x-13/2x-1

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Answer from: kayla941

To find the vertical asymptote, set the denominator equal to zero and solve for the variable x using algebra. The denominator is 2x-1 so if you set that equal to zero that would be 2x-1=0 ⇒ 2x=1 ⇒ $x= \frac{1}{2}$ is the vertical asymptote

To find the horizontal asymptote, there are three possibilities. Look at the degree of the numerator and denominator. Compare the degrees using the rules attached. Notice both the denominator and numerator have the same degree, so using the rule, the horizontal asymptote would be $y= \frac{a}{b}$ or in this case a=-8 and b=2 so the vertical asymptote is $y= \frac{-8}{2}$ or y=-4

Hope this helps
Graph all vertical and horizontal asymptotes of the function f(x)= -8x-13/2x-1

Graph all vertical and horizontal asymptotes of the function f(x)= -8x-13/2x-1

Answer from: annabanana1298

Refer the attached figure below.

Step-by-step explanation:

Given : Function $f(x)=\frac{-8x-13}{2x-1}$

To find : Graph all vertical and horizontal asymptotes of the function ?

Solution :

To find the vertical asymptote put the denominator of the function equal to zero.

The value of x from denominator is the vertical asymptote.

$f(x)=\frac{-8x-13}{2x-1}$

Denominator =0

2x-1=0

2x=1

$x=\frac{1}{2}$

So, The vertical asymptote is $x=\frac{1}{2}$

To find the horizontal asymptote we determine the degree of numerator and denominator.

Deg(num)=Deg(den)

So, $y=\frac{\text{leading coefficient of nr.}}{\text{leading coefficient of dr.}}$

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

y=-4

So, The horizontal asymptote is y=-4

Refer the attached figure below.

Answer from: Quest

$2^{2x+11}=3^{x-31}$

take the logarithm of both sides (the base of the log doesn't matter):

$\ln2^{2x+11}=\ln3^{x-31}$

apply the exponent property - this says that $\ln a^b=b\ln a$ :

$(2x+11)\ln2=(x-31)\ln3$

distribute the log terms on both sides:

$(2\ln2)x+11\ln2=(\ln3)x-31\ln3$

group like terms together:

$(2\ln2-\ln3)x=-11\ln2-31\ln3$

divide through both sides by the coefficient on $x$ :

$x=-\dfrac{11\ln2+31\ln3}{2\ln2-\ln3}$

you can do some additional rewriting here using the properties of the logarithm to simplify things if you like:

$x=-\dfrac{\ln2^{11}+\ln3^{31}}{\ln2^2-\ln3}$

$x=-\dfrac{\ln2^{11}3^{31}}{\ln\frac43}$

$x=-\log_{4/3}2^{11}3^{31}$

Answer from: Quest

b. 18

step-by-step explanation:

so the equation for this is:

ab^2=ac⋅ad

ab being the top length of 6 and ac being the length of 2 and ad being the length of x. let's plug into the equation:

6^2 = 2x

now lets just solve for x:

36 = 2x

18 = x

i hope that !

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Graph all vertical and horizontal asymptotes of the function f(x)= -8x-13/2x-1...

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