Description: UCI Math 1A/1B: Precalculus is designed to prepare students for a calculus course. This course is taught so that students will acquire a solid foundation in algebra and trigonometry. The course concentrates on the various functions that are important to the study of the calculus.
let's learn about the slope of a line
for example let's find the slope of the
line passing through these two points
here the slope of a line measures its
steepness and can be computed by the
following formula that the slope M is
the difference in the Y coordinates of
the two points which we call the rise
divided by the difference in the x
coordinates which we call the run so
let's let x1 y1 be our first point over
here this negative 4 negative 3 and X 2
y to be the second point namely 8
negative 9
so by this formula we have M the slope
is equal to y 2 which is negative 9 here
minus y 1 so minus a minus 3 be careful
/ x 2 which is 8 minus x1 which is
negative 4 which is equal to negative 9
plus 3 is negative 6 divided by 8 minus
a minus 4 is 12 which means that we just
found that this rise of a run down here
is equal to negative 6 over 12 so we can
think of negative 6 as the rise and 12
as the run and let's look at what this
means the run is positive 12 which means
from that point negative 4 negative 3
we're going to go to the right 12 and
then the rise is negative 6 so we go
down six to get to that other point
let's see what happens if we compute em
the other way in other words let's let
this point 8 negative 9 be x1 y1 instead
so if we let x1 y1 equal 8 negative 9 +
x2 y2 be equal to the other point
negative 4 negative 3 then by this
formula we get M is equal to y 2 which
is negative 3 minus y 1 so minus a minus
9 divided by X 2 which is negative 4
minus x1 which is 8 which is equal to 6
divided by negative 12 so now we can
think of our run as being negative 12
and our rise as being 6
that is from our first point 8 negative
9 our run is negative 12 s we go to the
left 12 and our rise is positive 6 so we
go up six units to get to the other
point so it doesn't matter which way we
label our points in either case we get
that our slope is equal to negative 6
over 12 which is negative one-half let's
look at another example let's find the
slope of this line that's graphed here
alright so this is the point negative 1
negative 2 and this is the point 2 3
again we're going to apply our slope
formula here that the slope is y2 minus
y1 divided by x2 minus x1 so let's let
this point be x1 y1 and this point be x2
y2 by our formula here m is equal to y2
which is three minus y1 which is
negative two divided by x2 which is two
minus x1 which is negative one or 5/3 so
we can think of our run as being
positive 3 and our rise as being 5 which
means starting at our first point we go
to the right three units and then up
five units to get to our second point
now we could have done it the other way
right we could have let this point here
BR x1 y1
and this point here BR x2 y2 so by our
formula then the slope would be equal to
y 2 which is negative 2 minus y1 which
is 3 divided by X 2 which is negative 1
minus x1 which is 2 or negative 5
divided by negative 3 so the run is
negative 3 rise is negative 5 that is
from our first point we go to the left 3
units and then down 5 units to get to
our second point but in either case we
get that the slope is equal to 5 thirds
and this is how you compute the slope of
a line thank you and we'll see you next
time
you