```Question 116018
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(-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(-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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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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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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