HELLO Mr. Tarrou. Today we are going to be talking parabolas. And I will see if i can
get this video done if fifteen minutes and if can't this will be two parts. Lets talk
about transformations. We learned about those not very long ago. I am just going to put
up here a very basic parabola. This is y equals x squared. Now do you remember the transformations
were like? Say if I wanted to take this parabola and shift it right two and down three units
and reflect the parabola like this. Remember what made those parabolas and the other seven
parent functions that we studied move around like that? Right, down, reflect, horizontal
flips and all that other good stuff. Well if you do, great and if you don't let me help
you out. This is going to be f of x or y. Now I moved this parabola to the right and
down. Do you remember how left and right movement showed up? That was when we had a plus or
minus inside the function itself. So our function is a power of two. We are going to have for
this movement to the right x minus two inside my function. What else did I do to the parent
function? I moved it down three. Where do we find vertical shifts at? That is the constant
at the end of your equation (usually), so minus three. This will take this parabola
and move it down three places. Now what caused the refection over the x axis when we studied
transformations of functions? That was when you equation started off with a negative value.
So this red equation now is the equation from the parent function which has been moved to
the right two and down three. The was the reflection about the x axis, the minus two
is the movement to the right, and the negative three at the end is our shift down. Well the
transformations chapter of our book... Of if you want to go back and watch another video...
if you are not in my class, helps us understand what the standard form of a parabola comes
from. The standard form for parabolas is f of x equals plus or minus a... That is possible
a reflection about the x axis. If a is larger than one, the graph is going to get pulled
vertically and will appear to be skinnier though it really is not. If a is less than
one and greater than zero, the graph has been compressed... from your transformations section.
x minus h squared plus k. This is the standard form of a parabola, which is one of the conic
sections that we are going to learn about today. When you look at this standard form,
you can tell a lot about the graph. You can tell where the vertex is. This vertex of the
parabola is at (2, -3). You can see that right here. It is two, we want to change that sign
when we take it out of parenthesis, negative three. Or, in standard form the vertex is
(h,k) So, hopefully that looks familiar now to the standard form getting tied into the
transformations section we covered a few weeks ago. What else is in here? All parabolas have
an axis of symmetry. A line that you can put down the middle of the parabola. We are only
going to do vertical parabolas today so that is going to be when x is squared. If y is
squared the parabolas are horizontal. We will study those later in the year. So if you have
an x squared term, you have a vertical line which is an axis of symmetry. Vertical lines
have an equation in the form of x equals a number. What is the x value that the axis
of symmetry is going to go through? It is going to go through the vertex, so it is going
to be x equals h. Or x equals whatever the x component is of your vertex. In this case,
this will be x equals positive two for the axis of symmetry. Um, there are other things
we can find that we are going to do in minute like x and y intercepts and also a parabola...
A graph has a relative maximum when it has a peak and then it falls on both sides. This
red parabola will also have a relative maximum, actually absolute maximum. The maximum will
be at two...negative three. We have been talking about domain and range as well this year.
The domain, what x values does this graph include? All of them from negative infinity
to positive infinity. What is the range? The range is negative three all the way down to
negative infinity. In case you need some help with that. Domain are you x's. And in some
class you may say all real numbers. In Precalc we would say the domain is negative infinity
to positive infinity... or in calculus. The range in our particular example, the red parabola...
well that is the y values and the largest y value is negative three and it goes down
from there. Thus the range is y is less than or equal to negative three. Interval notation
that my kids are using starts with the smallest y value which is negative infinity. The smallest
y value is negative infinity going up to a y value of negative three. The square bracket
indicates the value of negative three is actually used. There we go. A little tie in between
our old chapter about transformations... left, right horizontal movement, up and down movements,
vertical flips and how that shows you we get the standard form of a parabola. Now most
of our work today from here on out is going to be about working with the general form
of a parabola. So lets do that. We are going to do one example. Graph it and find solutions
and do a nice graph, not just a sketch like I have been doing with my students lately.
la la la...la la la... Ok. Now that is clean enough. Let take a look at this example of
a parabola. We got y equals negative two x plus three x squared minus four. Ok now this
parabola is not in what we would call general form. Because in general form you want your
exponents to count down as you go to the right. Mine don't. I have an exponent of one and
then I have a two. So I am going to rewrite this as y equals three x squared minus two
x minus four. And go into.... Let me use function notation. The general form of a parabola is
ax squared plus bx plus c. And we are going to graph this parabola in general form. Later
on this year we will be studying conic sections a lot and we will be using completing the
square process to put this general form into standard form. But, we are not doing completing
the square today and putting these into standard form. So what if you are given a parabola
and you are asked to graph the parabola in this form? Well step one is to find the x
of the vertex. That is found by doing negative b over 2a. Which is negative negative two
because it is a, b, and c. Here is a, b, and c. Over two a which is three. And this is
going to come out to be two over six or one third. Ok great! I am starting to understand
where the vertex is at. Now the y part of the vertex is found by taking this x value
and plugging it back into the function. So that is going to be three times one third
squared minus two times one third minus four. That is going to be three times one ninth
minus two thirds minus four. That little fraction work today is going to be one third minus
two thirds minus... Now I want to put all three of these numbers together and my four
has a denominator of one. So I am going to multiply the top and bottom by three so we
have a common denominator. That will be negative three, oops no it won't, it will be negative
twelve...yes minus twelve over three. So one minus two is negative one, negative one minus
twelve is negative thirteen over three. And this parabola has a vertex which is at one
third negative thirteen over three. Good. Now here is what I have got so far. I am going
to have to erase this by the way because I don't think I will fit all this on my board.
But uh... 1,2,3, 1,2,3 1,2,3 1,2,3,4 Lets see if this is enough to make our sketch.
Our vertex is at one third comma negative thirteen over three. How many threes are in
thirteen? Four leaving a remainder of one, so that is going to be negative four and one
third. A little less than negative four. We need to make another tick mark. One third
and a little bit past negative four. About right there is the vertex. Now, do you see
this leading coefficient? When the leading coefficient is positive our parabola is going
to open up. If this three was negative the parabola would open down. So I know that my
parabola is going to open up from here. What other points could we easily find to help
us graph this parabola and not just make a sketch. Lets take a look at the y intercept.
For a y intercept you want to let x equal zero. When you cross the y axis, what is x?
There is no horizontal movement so x will be zero. Now that will be fairly easy to do.
F of zero... I am switching to function notation... is going to be three times zero squared minus
two times zero minus four. So F of zero, my y intercept, is going to be negative four.
SO, I have another point right here. Shall I color code that? My y intercept is right
here. Ok...Good. Now I am going to need to erase some of this work to help us find the
x intercept because that is going to involve factoring or the quadratic formula. Now I
have shown you a little bit of factoring recently so I am going to remind you of how to do the
quadratic formula again. Which actually I have done that recently as well. But that
is what we are doing. Anyway our y intercept is when we let x equal zero. We are going
to find our x intercepts by letting y equal zero. ohhh... almost erased my problem. So.
I want to find the x intercepts of this. That means let y equal zero. So what I want to
do is solve for again the x intercept. Solve this equation. Three x squared minus two x
minus four equals zero. Alright! Lets see what this looks like. With the quadratic formula...
in case you don't remember what that is. The quadratic formula is x equals negative b plus
or minus the square root of b squared minus four a c all over two a. You will with me?
Alright! So if we are going to apply this quadratic formula to the quadratic equation
that is in general form. That means this is a which is three, b is going to be negative
two, and c is negative four. So that is going to be x equals negative b, watch the parenthesis
when you are plugging in these numbers... it helps, plus and minus the square root of
b squared minus four a, c, all over two a. That is going to be two plus or minus the
square root of four plus that is four and twelve is forty eight over six. That is going
to be two plus or minus the square root of fifty two over six. Now if you want to reduce
this and have your answer in exact form, fifty two is .... the square root of four times
the square root of thirteen over six. So this in exact form is two plus or minus two square
root of thirteen over six. haha....do you know where this is on the x axis? I don't
either! So if you type this into your calculator, or this into your calculator, you are going
to get your decimal approximation. I am going to cheat with my notes. It is negative point
eight seven and one point five. So this how you solve a quadratic equation using the quadratic
formula. I am pretty sure this could not have been done with factoring as well. Now... that
means that I have an x intercept of negative point eight seven. So at negative eight point
seven and at positive one point five... and voila we have our parabola. Now lets see here
Domain before I run out of time. The x's from negative infinity to positive infinity because
that parabola will open up forever. The range is going to be for my y values. My smallest
y value is negative four point... I erased it OH NO!!!... negative four point three three
to infinity. The axis of symmetry is a vertical line with an equation of x equals, I erased
everything too fast. Well whatever that x value was of the vertex, that is the axis
of symmetry. WOOO Go Do Your Homework. Thank you!!!