Question 116009
a)


<b>Description of transformation</b>:

Remember, {{{f(x)}}} is the same as {{{y}}}. So this means  {{{y=e^x}}}.

So when we say {{{e^x+3}}}, we're also saying {{{y+3}}} (replace {{{e^x}}} with y). So this means we're adding 3 to each y value which graphically shows us that we're shifting each y value up 3 units. 


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Answer:

So the transformation {{{g(x)=e^x+3}}} simply shifts the entire curve up 3 units.



Notice if we graph {{{f(x)}}} and {{{g(x)}}}, we get


{{{ graph( 500, 500, -10, 10, -10, 10, exp(x),exp(x)+3) }}}  Graph of {{{f(x)=e^x}}} (red) and  {{{h(x)=e^x+3}}} (green)


and we can visually verify the transformation


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<b>Horizontal Asymptote</b>:

Now if we found the asymptote of  {{{y=e^x}}}, we would find that the asymptote is {{{y=0}}}. Since we're translating each point on  {{{y=e^x}}} up 3 units, we're also translating the horizontal asymptote up 3 units. 


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Answer:

So the new horizontal asymptote is {{{y=3}}}.


Also, you can visually verify this answer by looking at the graph above



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<b>y-intercept in (x, y) form</b>:


If we let x=0 and plug it into {{{y=e^x}}}, we get


{{{y=e^0}}} Plug in x=0


{{{y=1}}} Raise e to the zeroth power to get one. Remember any number x to the zeroth power is always one (ie {{{x^0=1}}})


So for {{{f(x)}}} the y-intercept is (0,1)



Now if {{{g(x)}}} translates each y value up 3 units, then simply add 3 to the y-coordinate of the y-intercept to get


(0,1+3)---->(0,4)


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Answer:

So the new y-intercept is (0,4)


Once again, you can visually verify this answer by looking at the graph above



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b)


<b>Description of transformation</b>:


Looking at  {{{h(x)=e^(-x)}}}, notice how the exponent is negated. So let's see what affect this transformation has on  {{{f(x)=e^x}}}, 


 {{{h(x)=e^(-x)}}} Start with the given transformation 



 {{{h(-2)=e^(-(-2))}}} Plug in x=-2



 {{{h(-2)=e^(2)}}} Negate {{{-(-2)}}} to get 2



Notice if we plug in x=2 into {{{f(x)=e^x}}},  we get {{{f(2)=e^2}}}


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 {{{h(x)=e^(-x)}}} Start with the given transformation 



 {{{h(-1)=e^(-(-1))}}} Plug in x=-1



 {{{h(-1)=e^(1)}}} Negate {{{-(-1)}}} to get 1


 {{{h(-1)=e}}} Remove the exponent of 1


Notice if we plug in x=1 into {{{f(x)=e^x}}},  we get {{{f(1)=e^1=e}}}



So if we take the opposite of x (to get -x), and plug that into g(x), we'll get the same f(x) answer.



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Answer:

So what this does is simply reflect the entire graph over the y-axis


Notice if we graph {{{f(x)}}} and {{{g(x)}}}, we get


{{{ graph( 500, 500, -10, 10, -10, 10, exp(x),exp(-x)) }}}  Graph of {{{f(x)=e^x}}} (red) and  {{{h(x)=e^(-x)}}} (green)


and we can visually verify the transformation



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<b>Horizontal Asymptote</b>:


Since we reflected the graph with respect to the y-axis, the horizontal asymptote of  {{{h(x)=e^(-x)}}} is the same as  {{{f(x)=e^x}}} (you can see this from the graph above) 

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Answer:

So the horizontal asymptote is {{{y=0}}}


Note: you can visually verify this answer by looking at the graph above

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<b>y-intercept in (x, y) form</b>:


Since the line of symmetry between the two graphs is the line x=0 (ie the y axis), this means that the point that intersects with the y-axis is reflected to itself. So essentially the y-intercept does not change also. Once again, you can visually verify this using the graph above.


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Answer:

So the y-intercept is (0,1)


Once again, you can visually verify this answer by looking at the graph above