Question

What is the equation of the asymptote of the graph of y= 3 (4^x)

A. x=4
B. x= 0
C. y= 3
D. y= 0

Answer 1

This is an exponential function. What property do they all share? That they don't cross the x - axis. So if that is true, what would your asymptote be?

Answer 2

You need to find \[
\lim_{x\to\infty}y
\]and \[
\lim_{x\to-\infty} y
\]if you want to to find the horizontal asymptotes.

Answer 3

@mathstudent55 @mathmale @pooja195 @Luigi0210 @ParthKohli @dan815 @robtobey @Mehek14 @jigglypuff314 @jabez177 please help

Answer 4

One is this assumed to be a calculus question? Well, technically it is, but more it is asked to people who don't know about limits.

Answer 5

In any case, I will try to give you a similar example.

Answer 6

its algebra

Answer 7

\(\color{#000000}{ \displaystyle f(x)=7\cdot (2)^x }\)

I want to introduce to you, one important property:
\(\color{#000000}{ \displaystyle a^{-b}=\frac{1}{a^b} }\)
Keep this one in mind.

Now, you know that as \(x\) gets smaller and smaller (or, alternatively, \(x\to -\infty\), if you've seen that notation before). \(\color{red}{*}\) Let's try to evaluate the function at some very small values of \(x\) and see what happens to the function, the smaller \(x\) gets.

\(\color{#000000}{ \displaystyle f(-1)=7\cdot (2)^{-1}=7\cdot \frac{1}{2^1}=\frac{7}{2}=3.5 }\)

\(\color{#000000}{ \displaystyle f(-8)=7\cdot (2)^{-8}=7\cdot \frac{1}{2^8}=\frac{7}{256}\approx 0.02734375 }\)

\(\color{#000000}{ \displaystyle f(-20)=7\cdot (2)^{-20}=7\cdot \frac{1}{2^{20}}=\frac{7}{1048576}\approx 0.00000667 }\)

So you see where this is approaching, the smaller value you take?

Answer 8

i cannot see the number cuz it says ' math processing error' :(

Answer 9

Well, try refreshing please.

Answer 10

it doesnt work i refresh it 5 times and still doesnt work