BAM! Mr. Tarrou. Welcome to part two of three parts where we're going to be discussing the
fundamentals of finding the equation of tangent lines this time we're going to be finding the
equation of a tangent line that is parallel to a given line so we are not specifically told to find
a tangent line at a point of tangency a point that we're just given
that is on the function we are given a line that is not a part of our function that it's somewhere
out there on the cartesian plane and we want to find in this case maybe we don't just find one
tangent line that is parallel to the given one for the function g of x is equal to x cubed plus two
now there's lots of things we have to learn in calculus but the less you memorize without
reference to something else the better off you're going to be so this this question here is asking
for the equation asking for the equation of a line so if you've never seen one of these problems
there's no need to stress out just remember the fundamentals from algebra 1. what do you need to
write the equation of a line you need a point you need a slope and then you can put that into slope
intercept form or standard form if you for some reason wanted to start with that or my favorite
i start with the point slope form so that's we're going to do what is the slope that we need
what is the slope that we're going to use to write the equation of this line
well i know this is calculus and i know that there's a function here we're about to do some
calculus to it but this isn't the first time you were asked to find the equation of a line parallel
to some other line the difference between this question and calculus and maybe what you would
have in algebra 1 or algebra 2 is if i just i could just give you a point and say you know
give me the equation of a line through you know whatever this point that is parallel to this given
line and you knew you'd still know hopefully that all we have to do is put this equation into a form
that we can easily identify the slope and there's more than one way to do that but my favorite is
to go into slope-intercept form so let's do yellow for 1 here and we want to find again that slope
we're going to take the equation of our line 3x minus y
minus 4 is equal to 0. we're going to solve this for y so we're going to add both sides by y and we
have minus four is equal to y and if you want to we can write that as y is equal to three x minus
four just to make it really fit perfectly that y equals m x plus b so the given line has a slope
which is equal to three okay well we're going to use that in a minute but now we
need to think about you know what is the deal with tangent lines we have a
cubic function and i'm going to do a quick sketch of this situation here in a minute but i want to
make sure i don't run out of room but we have a cubic function that is you know going to be
changing it's not a linear function its slope its gradient is always changing throughout its domain
we need to figure out how to find the what the instantaneous rate of change of this
cubic function such that or when is that instantaneous rate of change equal to three
okay oh okay so we need to find or and work with the derivative of this
function because we need to find a point of tangency to which we're going to run that
line which has a slope of 3 through and giving you a little preview here we might be finding
more than one tangent line so we can't know that point of tangency that point that we're going to
use with the slope to write the equation of a line until we know where along the x axis
does that derivative does that instantaneous rate of change equal to three so enough talking
step two is and we'll do that in orange step two is we're going to we're still working
with slope basically but it's going to be working with the slope or the derivative
g prime of x we're going to take the derivative of g of x
which means we're going to take the derivative of the left and the right hand side
so we have g prime of x is equal to the derivative of a variable base and fixed power power rule
we're going to bring that power down so the 3 up front we're going to reduce that power by 1
and so we're looking for we're looking at the derivative of x cubed is 3x squared now we're
going to move on and take the derivative of the second term of this expression which is
the derivative of 2 well like y equals 2 is a horizontal line those those have slopes of 0 thus
that reminder that the derivative of a constant is equal to zero well now we have a function
three x squared which is telling us the slope of our original function g x cubed plus two now we
just need to find what is the x value such that the derivative the instantaneous rate of change
is equal for function g is equal to three so this is still on the vein of finding slope but um it's
like first we think of the slope of the line maybe 2 we're talking about the slope of that function
and so we're going to keep working with this derivative we have 3x squared is equal to 3.
we're going to divide both sides by 3 and get x squared is equal to 1. now we need to of
course get rid of that power of 2 by introducing a square root we're going to take the square root of
the square root function of both sides an even power of 2 and an even root of 2
the square root of x squared that acts as an absolute value function so really the absolute
the the answer to the square root of x squared is the absolute value of x which then is equal to 1.
your teacher and honestly i don't even show that very often it's just a reminder though
of like where is that plus or minus coming from when you take the square root of both sides
well it's from the square root of x squared acting as an absolute value function so we have when
x is either one or negative one those are two locations along the x-axis where our original
function g of x has a slope which is equal to three of course we found that by connecting and
setting equal the derivative of g to that slope of 3. well if there's two places along the x-axis
where our original function g or the derivative g prime of x is equal to three well then that means
we have we're going to have two points of tangency ultimately giving us two tangent lines that are
parallel to our given line of three x minus y minus four so let's find those point of tangencies
so step four we'll do that in blue or i guess well step three we're going to find those points
of tangency by well plugging an x value of 1 into the original function
so g of 1 is equal to 1 cubed plus 2 which is going to give us a point of 1 comma three
and then we have g of negative one negative one cubed plus two is going to be negative
one cubed is negative one plus two is one so we have another point of tangency of
negative one one and just to double check that i'm not doing some kind of silly
mistake okay now we're going to use each of those point of tangencies along with
that that slope that we want to work with that parallel slope from the line that slope of three
and find our two tangent lines so tangent line one we'll go we'll keep that in i
guess we'll keep both in blue y minus y sub 2 y minus 3 is equal to 3 times x minus 1. ap
is never going to be happy with the final answer in point-slope form they're always going to want
them in slope-intercept form so we're looking at y minus 3 is equal to 3x minus 3 add 3 to both sides
and we have y is equal to 3 x working off on our second one over here we have
y minus y sub 1 or y minus 1 is equal to 3 times x minus negative 1. so we're looking at y minus
1 is equal to 3x plus 3 and y is equal to three x plus four so we have two tangent lines we have
that idea that we see this question or ask for the equation of a tangent line maybe the information
is given in a slightly new format for us but don't forget your basics we found the slope
of the line we found the slope of function g in the form of g prime we set that equal
to three we found two values of x where the instantaneous rate of change or the gradient
of g of x was equal to three and that led us to our two lines we have tangent line one
y is equal to three x and tangent line two
which is that y is equal to 3x plus 4. so for a quick visual here we have a standard cubic
function a parent function that you learned in algebra 2 and precalculus reminded of in
precalculus we have a cubic function that's been shifted up 2. so instead of passing through 0
0 we're passing through 0 2 and there's no vertical stretch on this cubic function there's
no leading coefficient other than one so it shifts around a bit but over if we're going
to go over one and up one and over one and down to the left one and down one if there was a two
out in front of this cubic i would go over one and up 2 because 1 squared is 1 but then you double it
and our cubic function looks something like this this is just going to be a sketch
a function that was an odd function the original cubic function is symmetric to the origin
once we shift it up two units of course it's not or anywhere off of the origin if you will we have
tangent line one we'll do that in blue where we have y is equal to three x with a y intercept
of of course zero so we're going to go up three and over one and we've got this
line which shows that i probably curved my didn't curve my cubic function enough but
we have a tangent line here coming and intersecting at point one three
and let me grab a yellow chalk here or yep there we go the yellow one and our second tangent line
is y is equal to 3x plus four same slope just a vertical shift of four so up here to one two three
four we have a slope of three so up three and over one maybe down three and over one and my cubic
functions kind of showing like this is a secant line so let me fix that of course if you use a
desmos or graphing calculator it wouldn't be uh having to fix your drawing but it's always a
good idea when you're doing calculus it's really important that you understand the relationship
and the reason why i'm doing this is it's it's really important for you through the course
to and to get that deep understanding of what all these functions do and how they relate to
each other is graph them we have a function with the tangent lines they should be just
touching and looking like they're basically going in the same direction or parallel to that function
at that point of tangency where we just have that one instantaneous rate of change and that slope is
constantly changing if you have a function and you find its derivative which i now erased would be
a good idea to to take that cubic function and then right on top of that maybe y sub 2 put in
the derivative so you can see how the derivative function and the original function also relate
to each other but that is the end of our example that is example number two or a second form type
of question where you'll be finding equations of tangent lines in the third video we're going to be
finding the equation of a tangent line that goes through a point that is not on the curve so it's
going to be a little bit more involved because it won't be so easy to identify what the slope
of that tangent line what it needs to be but for now i'm done i'm mr tarrou bam go do your homework
you