+0

# What is the inverse of h? h(x) = 6x + 1

0
252
2

What is the inverse of h?
h(x) = 6x + 1

A)h-1(x) = 6x - 1

B)h-1(x) = 6x + 1

C)h-1(x) = x/6 -1

D)h-1(x) = 1/6(x - 1)

What are the vertical asymptotes of R(x) = 3x - 3/x2 - 4?

A)x = 0

B)x = 1, and x = -1

C)x = 2, and x = -2

D)x = 3, and x = -3

Guest Aug 15, 2017
Sort:

#1
+80941
+1

h(x)  = 6x + 1     we can write

y  =  6x + 1         the object  is to get x by itself and then "swap" x and y

Subtract 1 from both sides

y - 1  = 6x    divide both sides by 6

[ y - 1 ] / 6   =  x       "swap x and y

[x - 1 ] / 6  = y         for y, write  h-1(x)

[ x - 1 ] / 6  =    h-1 (x)         and this is the inverse

What are the vertical asymptotes of R(x) = 3x - 3/(x2) - 4?

As shown here, the vertcal asymptote occurs at x  = 0

https://www.desmos.com/calculator/uyregpotef

CPhill  Aug 15, 2017
edited by CPhill  Aug 15, 2017
#2
+1602
+1

In the second question, Cphill interpreted it correctly, but if you meant $$R(x)=\frac{3x-3}{x^2-4}$$ or $$R(x)=3x-\frac{3}{x^2-4}$$, the answer is different. Let's think about what we have to do to figure out the vertical asymptote.

In both cases, we have rational fractions. Of course, we can't have a denominator of 0. This means that if we plug in a value for x that results in a denominator that equals 0, then it is officially outside of the domain. Let's figure out when x^2-4=0.

 $$x^2-4=0$$ You might notice that x^2-4 is a difference of 2 squares, but we don't need to take advantage of this, actually. This is because we have no b-term. Add4 to both sides. $$x^2=4$$ Take the square root of both sides. $$x=\pm2$$

This means that the vertical asymptote is at $$x=\pm2$$, which corresponds to the answer choice of C.

TheXSquaredFactor  Aug 15, 2017

### 21 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details