Question 4873
graphing....learn the basic shapes first then apply that knowledge to more complicated ones. So {{{y=2^x}}} (or any exponential function) looks like:


{{{graph(200,200,-4,5,-6,40, 2^x)}}}


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(x) = 2^4x+4}}}? First thing is USE BRACKETS!!!! because i do not know if you mean {{{f(x) = 2^(4x+4)}}} or {{{f(x) = 2^(4x)+4}}}. So i shall do both!


1. {{{f(x) = 2^(4x+4)}}}


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^(4x+4)}}} --> {{{y = 2^(4(0)+4)}}} --> {{{y = 2^4}}} --> y=16.


{{{graph(200,200,-2,1,-3,300, 2^(4x+4))}}}


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


2. {{{f(x) = 2^(4x)+4}}}


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(x) = 2^(4x)+4}}} --> {{{f(0) = y = 2^(0)+4}}} --> {{{y = 1+4}}} --> y=5


{{{graph(200,200,-2,1,-3,30, 2^(4x)+4)}}}


Hope this all helps. Learn the following basic shapes:


{{{f(x) = x^2}}}
{{{f(x) = x^3}}}
{{{f(x) = x^3 + other terms}}}
{{{f(x) = 1/x}}}
{{{f(x) = 2^x}}}
{{{f(x) = log(x)}}}


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


jon