In the last video about quadratic
functions we talked about functions that
looked a little something like this.
You take the basic X-squared graphed you
shift it right and left and you shift it
up and down. This form of a quadratic
function very easily gives you the
vertex of the function it's just HK and
you can even multiply this function by
some number and that number will stretch
and squish the function vertically and
it can even flip it over the X-axis.
However not every quadratic function
comes to us in this form. A lot of times
we have a function of the form AX
squared plus BX plus C, for example if a
is 2B is negative 5 and C is negative 3
our quadratic function might look
something like that and the main
question I want to answer in this video
is how do we find properties of this
function so that we can graph it. In
other words how do we find its
X-intercept its Y-intercept its vertex.
In order to answer one of those
questions we're gonna have to use this
thing called a quadratic formula. Okay so
that's where we're heading let's get
started. Let's start by finding the
Y-intercept of this graph which we
always just find by replacing X with zero.
If we do that we get F of zero equals
negative 3 which means we have a Y-intercept of zero
negative 3 and we'll start by graphing
that point right there.
Well how about X-intercepts, well to find
X-intercepts we replace Y with zero or we
replace the function value with zero and we solve for X. In this specific example we
can factor the right-hand side using
whatever method you like. That is how to
factor that appropriately you can split
this up into two different problems and
we get two X-intercepts. We get negative
1/2 zero and we get 3 zero. Okay well things seem
to be moving along pretty well one more
thing we might want to know about for
this quadratic function is where is its
vertex. The trick that we can borrow from
calculus is the X value of a vertex of a
quadratic function in this form is
always negative B over 2A and in this
example that means negative negative 5
over 2 times 2. The negative negative
becomes a positive and we get that the X
value of our vertex is 5/4ths. Well X equals
5/4ths so that's like a little bit bigger
than 1 so that's gonna be somewhere
along this line right here, but how do we
find that the Y value is for this vertex.
Well what we have to do is we just have to
plug this X value into the
function figure out what Y is
unfortunately that means dealing with
some fractions. The notation for this is
F of 5/4ths, if we square this fraction and
then multiply it by 2 we are getting 50
over 16. If we multiply 5 by 5/4ths we
get 25/4ths we leave the minus 3 here.
Now if we find a common denominator and
add these 3 things together we get
we get, negative 98 over 16 which
actually does reduce but it's
approximately negative 6.125. So
somewhere down here is our vertex in
it's 5/4ths negative let's see 98/16ths reduces to 49/8ths. Which is a
little bit less than negative 6. Okay great
so we can actually see what our parabola
is basically going to look like.
Beautiful drawing and we're pretty happy
with this problem. If we have a quadratic
of the form AX squared plus BX plus C we
can find the Y-intercept really easily
by plugging in X equals zero. We can find
the X-intercepts by plugging in Y equals
zero and we can find the vertex by using
this trick negative B over 2A it gives
us the X-value of our vertex. So that's
almost always gonna work. The problem is
this step right here. In this problem
that I made up we were able to factor
this right-hand side but that's not
always gonna be the case. What if we
can't factor this quadratic on the right
side of the equation to find our zeros.
We want to answer that question because
it's actually really easy to come up
with the quadratic that doesn't factor.
For example 2X squared plus 3X plus 4
equals zero. If we want to solve that for X
factoring just isn't going to work. So we
have this thing called the quadratic
formula that allows us to solve
equations of the form zero equals AX squared
plus BX plus C. The solution is given to
us in this somewhat complicated looking
formula let's use this quadratic formula
to solve for X in this equation right
here. The first thing you want to notice
is that our A value is 2, our B value is
3 and our C value is 4. Now we just have
to substitute A equals 2 to B equals 3 and C
equals 4 into the quadratic formula and
do our order of operations properly.
That's gonna look like this, I like to
keep my values that I'm substituting
into formulas in parentheses, just to
make sure I don't mess anything up. If we
do our order of operations, that means we
have to simplify this stuff under the
square root first. Well this is giving us
9 minus 32 under the square root. It's
giving us a negative value on
square root which we know from previous
experience is not a real number.
So for now we're just going to say no
real solution. If we had a function 2X
squared plus 3X plus 4 this operation
would be finding that function zeros and
we would have just discovered that this
function has no zeros it doesn't cross
the X-axis. Okay so let me alter the
problem so that we'll get an answer
that's just a little bit more satisfying.
What if we want to find the zeros of 2X
squared plus 3X minus 4. Well what we do
is replace the function with zero. We might
try to factor this right-hand side but
we can't. So we have to use the quadratic
formula and we're gonna use this
quadratic formula with A equals 2, B
equals 3 and C equals negative 4 this
time. I'll make the substitutions I'm
going to simplify what's under the
square root. You'll notice we get 9 minus a negative
32. That gives us 9 plus 32 which is now
a positive number under our square root
and we have two zeros negative 3 plus
the square root of 41 over 4 and
negative 3 minus the square root of 41
over 4. If we plug these into our
calculator we get approximately, 0.85 is one answer and approximately negative
2.35 is our other answer. 
If we wanted to make a little graph of
this 0.85 is a little bit less than one,
negative 2.35 is in
between negative 2 and negative 3.
Finding a Y-intercept is usually very
easy so I'll plug in X equals zero and
we get F of zero equals negative 4. So
we have a point down here at zero
negative 4 and a basic sketch of our
parabola will look something like this.
If you're curious as to what this vertex
is, ah figure it out use the formula X
equals negative B over 2A to find the
X-value then plug that X value into the
function to find the Y value. Okay
obviously we'll work more with quadratic
functions in the next class and also in
the next video so I'll see you there.
