V1 Math 141 – Finals Review

V1 Math 141 – Finals Review


This is video one for math 141 finals
review. Given two points we’re going to create a linear function. I’m going to
start off with thinking about which point is point number one and point
number two, and I’m going to label my point X sub 1, Y sub 1 and X sub 2 y sub
2. The first thing you want to do when you create a linear function is to find
the slope. So I’m going to take the change in Y, and that is over the change
in X. For this particular two points my Y sub 2 is the number 2, and my Y sub 1 is
3. My X sub 2 is 3 and my Y sub 2 is a negative 2. When we go through this
process 2 minus 3 is minus 1 3 minus a negative 2 is the same as 3 plus 2. So we
end up with a 5 in the denominator. So negative 1/5 is my slope. Once I have the
slope I choose one of my points as my X&Y that I’m going to replace in the
equation. Y equals MX plus B I now have a Y value I’m going to use, an x value and
the slope. My Y value is 2 my slope is negative 1/5. My x value is a positive, 3
and now I can solve for B negative 1/3 times 3 is a negative 3/5. To solve for B I have to add three
fifths to both sides, but three fifths doesn’t add to two very nicely, so I have
to change that so it has the same denominator. 2 is the same as 10 fifths.
When I add three-fifths to both sides that isolates might b, and I’m left with
13 fifths as my B. My final equation we’re going to call this a function so
we’re going to use f of X for this function. f of X is therefore equal to my
slope which is my negative 1/5th. X plus my slope 13 fifths. Now if you’re asked to
graph this you, can take the points negative 2/3, negative 2/3, there’s one
point. The other point is positive 3/2, positive 3/2, and with a straightedge you
would connect these two points. So that would be the equation- finding a linear
equation using two points. If you are then asked to find the inverse function,
the inverse function would be f to the negative 1 of X, you would want to switch
your x and y values. So instead of our equation here, f of X would be our Y and
if we want to find the inverse, we would make our Y value x, and our x value
would be Y. So this function would end up being, x
equals negative one-fifth y plus 13 over 15. Now I want to isolate my Y to get
this inverse function. So I’m going to subtract 13 fifths from the X. Sorry that
should have been a five instead of a 15 and take that away there. So I’m
subtracting 13 fifths from both sides that equals a negative one-fifth Y, and
then I’m going to multiply both sides by negative five, and so that would get rid
of the negative one-fifth that would make it a positive Y over here. Negative
5 X and this would be these 5s would cancel, the negative would cancel and I’d
have plus 13. So this would be our inverse function negative 5 X plus 13. That would be the inverse of this function here. To find the equation to
find it on the graph we would plot the points in the opposite order. Our X’s
become our Y’s and our Y’s become our X’s. So instead of a negative 2 as our X
the 3 is our X and the negative 2 becomes our Y. Here the 2 becomes our X
and the 3 becomes our Y. So when we plot these points we have positive 3 negative
2, and we have positive 2 positive 3. We would connect these for the straight
line, and this would be your inverse function.
So this would be f to the negative one of X, while our- whereas this is our f of
X. The domains and ranges are positive and negative infinity, that doesn’t
change. They’re both straight lines they go to infinity in both directions, but
these are the steps for finding an equation given two points and then going
through the process of finding the inverse function.

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