let's talk about transformations of exponential
graphs the first type of transformation that we should talk about our horizontal and vertical
shifts these are the easiest to understand recall that the graph of f of X plus h quantity
plus K is the graph of f of X shifted left by H and up by K the thing to remember here is
that if H is negative then that's a shift to the right if H is positive that the shift
to the left and similarly if K is positive to shift up okay is negative sit shift down so
in this particular context with exponential graphs if our function G of X is equal to B to
the X plus h plus K well that means we're going to shift up by K which means that we have a
horizontal asymptote at y equals K and then the other thing we can do is plot a convenient
point so if if X is minus H well when X is minus H this exponent becomes zero so you just get b
to the 0 which is 1 plus k so the y value there is just k plus 1 so we know that the point minus
h k plus 1 is on the graph and it's best not to try to memorize this but just to remember that
a convenient point is when the exponent is 0 that's an easy point to compute and so you
can always just figure out what this point is by plugging in the appropriate value of x and
finding what the value of y is at that point and just note that this is the same as the point 0
1 after it's been shifted left by H and up by K and then another point once you can get one more
point just by thinking about what happens when X is minus h plus 1 so when X is minus h plus one
well then now the exponent becomes minus h plus one plus h which is just one so we get B to the
one which is B plus K is B plus K again you don't have to memorize this point just remember that
that's another convenient point to compute when the exponent is one you just get B to the 1 plus
K and that's the same as the point 1 B after it's been shifted left by H and up by K so these are
just too convenient points on our original graph after they've been shifted left by H and up by
K so let's look at an example we're asked to find the graph of the function f of X equals 2
to the X plus 3 minus 1 so to start with let's think about what our ships are right we've got X
plus 3 so our H is three and we're subtracting one from its 0 RK is minus 1 that's a vertical shift
of minus 1 or 1 unit down and a horizontal shift of 3 which is three units to the left because H
is positive so again we can go about this in a few ways let's start by figuring out using this
to find our horizontal asymptote because our normal horizontal asymptote is just the x axis
y equals zero now it's going to be shifted down by one so our horizontal asymptote is going to
be at y equals minus 1 now let's just plot some convenient points so again we could try to
memorize them but it's easier just to figure out what they're going to be when X is minus 3
that's a convenient point because it makes the exponent zero so when x equals minus 3 what do we
get we get 2 to the minus 3 plus 3 which is zero minus one which is one minus one which is zero
so we get the point minus 30 is on our graph so let's plot that minus 3 is 0 is right there
and now another convenient point would be when x equals minus 2 when X is minus to the exponent
becomes 1 so x equals minus 2 gives us 2 to the minus 2 plus 3 which is one minus one so we get
2 minus 1 which is 1 so that tells us the point minus 2 comma 1 is on our graph so let's plot that
point minus 2 1 is right there and now again we know the general shape of our exponential curve
and now we have two points on it and we have the horizontal asymptote if you wanted to plot one
more point just for convenience we could try x equals 0 get our y-intercept and that would give
us 8 2 to the 3 which is 8 minus 1 which is seven so just one more point for our graph sake now we
can see that this is our usual exponential curve just going through these two points and then up
through there so that's roughly what our graph should look like and there's our actual curve
you can see that our rough approximation was pretty close right we got the right point minus 30
minus 20 and then 07 and we have that horizontal asymptote at y equals minus 1 so that's a pretty
good graph we had a pretty good approximation of the graph now let's talk about vertical stretches
remember that for a vertical stretch the graph of a times f of X is the graph of f of X stretched
vertically by a factor of a just remember that when a is greater than one we get up a vertical
stretch a true stretching where it gets pulled upwards or downwards and when a is between zero
and one it becomes compressed it sort of shrinks a little bit so just good to keep that in mind so
in our particular context of exponential graphs we have G of x equals a times B to the X well the
horizontal asymptote y equals 0 does not change because when you vertically stretch something if
it has a horizontal asymptote at 0 that doesn't change and then we can plot some convenient
points so the point 0 a is what you get when you plug in x equals 0 into this function you get
a times b to the 0 which is a times 1 which is just a so that's the point 0 1 after it's been
stretched vertically by a factor of a and then if you want another point we can take the point
1 comma a times B and that's just what you get when you plug in x equals 1 into our function
you get a times B to the one which is a times B again this is just the point 1b after it's been
stretched vertically by a factor of a so with that horizontal asymptote and two points that's all we
really need to get a good idea of what our graph looks like to see a demonstration of a vertical
stretch let's just take a look at our graph of our function f of X equals 1 times 3 to the X or
just 3 to the X and think about what happens when we change our a our leading coefficient so for
instance here's our graph as is where we notice that you know we go through the point 13 we go
through the point 0 1 what happens when we change a 22 well when a is to that stretches vertically
so now we go through the point 0 2 we go through the point 16 right that's 2 times 3 so when X
is 1 we go through the point 6 16 and similarly we can stretch it more if we stretched if a was
3 we get vertically stretched even more is for we go to the point 0 4 and so on so this shows
us what happens when we vertically stretch our exponential graph the horizontal asymptote state
is the same at y equals 0 we just go through a different y-intercept and we get stretched
vertically and just as an extra twist what would happen if we were to say stretch and shift
vertically so say we added one to it while we can still just use our same property say if we were at
3 times 3 to the X plus 1 we would be vertically shifted by one now our horizontal asymptote is
y equals 1 but everything else stays the same we'd still have a vertical stretch by a factor
of 3 and then we just have to shift by 1 so as an example let's find the graph of the function f
of X equals 2 times the quantity 3 to the X plus 1 so let's start by noticing that because of this
plus 1 we're going to be shifted vertically by one so let's start by thinking about what that
means well remember 2 times 3 to the X that has a horizontal asymptote at y equals 0 but then when
we add one that horizontal asymptote is going to be at y equals 1 so we can start by sketching that
in and now let's just plot a few key points so we don't have to memorize this we can just think
about what happens when X is a convenient value so for instance when x equals 0 that's convenient
because then the exponent is 0 we just get one here so we get 2 times 3 to the 0 plus 1 again 3
to the 0 is 1 so we just get 2 plus 1 which is 3 so that gives us the point 0 comma 3 on our graph
so 03 this is where your convenient value what if x equals 1 well if x equals 1 we get 2 times 3
to the 1 plus 1 2 times 3 is 6 plus 1 is 7 so that tells us the point 1 comma 7 is on the graph
so 17 and with those two points in our horizontal asymptote that's enough to really get a pretty
good sketch of this graph we know we're going to approach the horizontal asymptote like that
and then go vertical to the right so that's a rough idea of what the graph looks like and if
we try we look at the actual graph that's what the actual graph looks like so we were just a
little bit off but that's that was a good rough estimate of what the graph looked like and here
it is without our sketch on there and then we can see again horizontal asymptote it's at y equals 1
I go through the point zero three and 17 like we said now let's talk about more transformations of
graphs let's talk about reflections in the context of exponential graphs remember that the graph of
F of minus X is the graph of f of X reflected over the y-axis because every x value has been reversed
which gives us a reflection over the y-axis and then similarly the graph of minus f of X is the
graph of f of X reflected over the x-axis because now we take every value of our function and we
take the negative of it so that reverses or it takes the negative of all of the Y values and that
causes a reflection over the x axis let's look a little demo of some of these reflections of our
exponential functions so for instance here we have the function y equals 2 to the X and now let's see
what happens when we reflect it in various ways so start with if we take y equals 2 to the minus X
shown here in red that's a reflection over the y-axis so we've taken the negative of all of our
X values which reflects the graph over the y-axis so if we reset that and now if we look at what
happens when we take y equals minus 2 to the X well when y is minus 2 to the X this reflects the
graph over the x-axis all the Y values we've taken the negatives of all of them and that reflects us
over the x-axis and finally if we were to reflect both well that gives us minus 2 to the minus
X and what happens when we do that well it's we first reflect over the y-axis and then over
the x-axis or we can think of it as reflecting over the x-axis and then over the y-axis either
way we get this very nice symmetrical looking graph that's now been reflected both over the
y axis and the x axis let's look at an example let's find the graph of the function f of X
equals minus 2 times the quantity 3 to the minus X so here we have to be a little careful
because we're not only reflecting we're also stretching but that doesn't make things too
much different than what we've already seen so to begin with let's just think about what the
general shape of this is going to be because we have a minus in front and a minus X we know
that our normal curve which looks something like this is going to be reflected over the
y-axis and over the x-axis so it's going to be somewhere down here so let's just think
about exactly what that's going to look like we think about a few points that are going to
be on this curve well when let's just take some convenient values so when x equals 0 we get
minus 2 times 3 to the minus 0 so 3 to the 0 which is minus 2 so we know that the point 0
minus 2 is on our graph what else do we have well when X is 1 that'll be it sorry when X is
minus one that'll be a convenient value when X is minus 1 this becomes 3 to the positive
one so we get minus 2 3 to the minus minus 1 that's 3 to the 1 so we get minus 2 times 3
which is minus 6 so we have the point minus one comma minus 6 and now with those two points
and we know that our horizontal asymptote hasn't changed it's going to be the x-axis we can now
get a rough estimate of this plot it's going to look something like that so this is what our
curve looks like roughly so we can see what the actual curve looks like there's what it looks
like actually so are our estimate was pretty close