SOLUTION: I am completely stuck on this problem. Help! Graph the fuction: {{{f (x)= 2^4x+4}}} {{{g (x)= 3 *log(2,(x-7))}}}

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I am completely stuck on this problem. Help! Graph the fuction: {{{f (x)= 2^4x+4}}} {{{g (x)= 3 *log(2,(x-7))}}}       Log On


   



Question 4873: I am completely stuck on this problem. Help!
Graph the fuction:

f+%28x%29=+2%5E4x%2B4
g+%28x%29=+3+%2Alog%282%2C%28x-7%29%29

Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
graphing....learn the basic shapes first then apply that knowledge to more complicated ones. So y=2%5Ex (or any exponential function) looks like:

graph%28200%2C200%2C-4%2C5%2C-6%2C40%2C+2%5Ex%29

Note 2 things:
1. it does not go below the x-axis, since there is no value of x that can make y negative. In fact as x becomes more negative, the curve approaches the x-axis, ie approaches y=0. This "approaching, but never quite reaching" characteristic is called the asymptote.
2. the curve rises rapidly to infinity as x increases.

So, what about f%28x%29+=+2%5E4x%2B4? First thing is USE BRACKETS!!!! because i do not know if you mean f%28x%29+=+2%5E%284x%2B4%29 or f%28x%29+=+2%5E%284x%29%2B4. So i shall do both!

1. f%28x%29+=+2%5E%284x%2B4%29

we know the rough shape. Just need to know the asymptote and any values where the curve crosses the axes.

Any value of x gives a whole range of possible values of 4x+4. All of these, as powers of 2 will never give a negative answer for y. So the curve, again, does not go below the x-axis...ie the asymptote is y=0.
Voila, all we need now is find where the curve crosses the y=axis (ie where x=0)... y+=+2%5E%284x%2B4%29 --> y+=+2%5E%284%280%29%2B4%29 --> y+=+2%5E4 --> y=16.

graph%28200%2C200%2C-2%2C1%2C-3%2C300%2C+2%5E%284x%2B4%29%29

see how this is exactly the same type of curve but the axes scales are hugely different.

2. f%28x%29+=+2%5E%284x%29%2B4

Again, we know the rough shape..any value of x will never give a negative answer, so never below y=0. However, we then add 4 to all answers, so our asymptote is now no longer y=0, but y=4.

Find where the curve crosses the y-axis (where x=0)...f%28x%29+=+2%5E%284x%29%2B4 --> f%280%29+=+y+=+2%5E%280%29%2B4 --> y+=+1%2B4 --> y=5

graph%28200%2C200%2C-2%2C1%2C-3%2C30%2C+2%5E%284x%29%2B4%29

Hope this all helps. Learn the following basic shapes:

f%28x%29+=+x%5E2
f%28x%29+=+x%5E3
f%28x%29+=+x%5E3+%2B+other+terms
f%28x%29+=+1%2Fx
f%28x%29+=+2%5Ex
f%28x%29+=+log%28x%29

More complex versions of each type have their own characters but at least it is a start :-)

jon