How do you graph $f(x)=4(3/4)^(x+1)-5$ and state the domain and range?

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Answer 1 · 2017-11-08T19:36:50

See below.

The y axis intercept occurs where $x=0$ , so:

$4(3/4)^(0+1)-5=-2$ coordinate $( 0 , -2)$

There are no restrictions on $x$ so:

Domain is ${ x in RR }$

$lim_(x->oo)4(3/4)^(x+1)-5=-5$ ( y = -5 is a horizontal asymptote )

For $x< -1$

$4(3/4)^(x+1)-5$ becomes:

$4/((3/4)^(x+1))-5$

as $x->oo$ , $(3/4)^(x+1)->0$

So:

$lim_(x->-oo)4(3/4)^(x+1)-5=oo$

Range is:

${y in RR: -5 < y < oo}$

Graph:

enter image source here

How do you graph f(x)=4(3/4)^(x+1)-5f(x)=4(3/4)^(x+1)-5 and state the domain and range?