### Video Transcript

The following graph shows the
function π sub one of π₯ is equal to two to the power of negative π₯. Use this graph and plot the
function π sub two of π₯ is equal to π₯ plus three to find the solution set of the
equation two to the power of negative π₯ is equal to π₯ plus three.

In this question, weβre given two
functions π sub one of π₯ and π sub two of π₯, and weβre given a graph of the
function π¦ is equal to π sub one of π₯. Weβre asked to find the solution
set of an equation. And since π sub one of π₯ is equal
to the left-hand side of this equation and π sub two of π₯ is equal to the
right-hand side of this equation, the equation is π sub one of π₯ equals π sub two
of π₯. We can solve this equation
graphically. Any solution to this equation will
be a point of intersection between the curve π¦ is equal to π sub one of π₯ and the
line π¦ is equal to π sub two of to π₯. Because the point of intersection
would have the same π¦-coordinate and the π¦- coordinate is the output of the
function for the given π₯ coordinator, which means the outputs of the function would
be the same, so our equation would be solved.

We need to sketch the curve π¦ is
equal to π₯ plus three. First, we note that its
π¦-intercept will be at three. We can also find its π₯-intercept
by substituting π¦ is equal to zero. Solving this, we get that π₯ is
equal to negative three. We can then plot our line. Its π¦-intercept is at three, and
its π₯-intercept is at negative three. This then allows us to plot our
line. We just connect the π¦- and
π₯-intercept with a straight line. Then, the only point of
intersection between our line and our curve will be the only solution to our
equation. We can read off its π₯-coordinate;
its π₯-coordinate is negative one.

Then, since the question ask us to
write this as a solution set, weβll write this as the set containing negative
one. Therefore, we were able to show the
solution set of the equation two to the power of negative π₯ is equal to π₯ plus
three is just the set containing negative one.