Non-linear systems of equations 1 Video Lecture | Mathematics for Digital SAT

Solve the system of equations
by graphing.
Check your solution
algebraically.
Let's graph each of these, and
let's start-- let me find a
nice dark color to
graph these with.
Let me graph this top equation
in blue, this parabola.
The first thing to think about
is this going to be an upward
opening-- one, how did I
know it's a parabola?
That's because it's a quadratic
function: we have an
x squared term, a second
degree term, here.
Then we have to think about:
it is going to be upward
opening, or downward
opening parabola?
You see that it's a negative
coefficient in front of the x
squared, so it's going to be a
downward opening parabola.
What is going to be
its maximum point?
Let's think about that
for a second.
This whole term right here is
always going to be negative,
or it's always going
to be non-positive.
x squared will be non-negative
when you multiply it by a
negative, so it's going
to be non-positive.
So, the highest value that this
thing can take on is when
x is going to be equal to 0--
the vertex of this parabola is
when x is equal to 0,
and y is equal to 6.
So, x is equal to 0, and
y is 1, 2, 3, 4, 5, 6.
So that right there is the
highest point of our parabola.
Then, if we want, we can a
graph a couple of other
points, just to see
what happens.
So let's see what happens when
x is equal to-- let me just
draw a little table here--
2, what is y?
It's negative x squared
plus 6.
So when x is 2, what is y?
You have 2 squared, which is
4, but you have negative 2
squared, so it's negative 4
plus 6-- it is equal to 2.
It's the same thing when
x is negative 2.
You put negative 2 there, you
square it, then you have
positive 4, but you have a
negative there, so it's
negative 4 plus 6 is 2.
You have both of those points
there, so 2 comma 2, and then
you have a negative 2 comma 2.
If I were to graph it, Let's try
it with 3, as well-- if we
put a 3 there, 3 squared is 9.
It then becomes a negative 9
plus 3, it becomes negative 3,
and negative 3 will also
become a negative 3.
Negative 3 squared is positive
9, you have a negative out
front, it becomes negative 9
plus 6, which is negative 3.
You have negative 3, negative
3, and then you have 3,
negative 3.
So those are all good points.
Now we can graph our parabola.
Our parabola will look
something-- I was doing well
until that second part --like
that, and let me just do the
second part.
That second part is hard to
draw-- let me do it from here.
It looks something like that.
We connect to this dot
right here, and then
let me connect this.
So that it looks something
like that.
That's what our parabola looks
like, and obviously it keeps
going down in that direction.
So that's that first graph.
Let's graph this second one
over here: y is equal to
negative 2x minus 2.
This is just going
to be a line.
It's a linear equation, and the
highest degree here is 1.
Our y-intercept is negative
2, so 0, 1, 2.
Our y-intercept is negative 2.
Our slope is negative 2.
If we move 1 in the x direction,
we're going to go 2
in the y-direction, and if we
move 2 in the x direction,
we're going to move down
4 in the y direction.
If we move back 2, we're going
to move up 2 in the y
direction, and it looks like we
found one of our points of
intersection.
Let's just draw that line, so
that line will look something
like-- It's hard for my hand to
draw that, but let me try
as best as I can.
This is the hardest part.
It will look something like
that right there.
The question is, where
do they intersect?
One point of intersection does
immediately pop out at us,
because they asked us to
do it graphically.
That point right there, which
is the point negative 2, 2.
It seems to pop out at
us, so this is the
point negative 2, 2.
Let's see if that makes sense.
When you have the point negative
2, when you put x is
equal to negative 2 here,
negative 2 times negative 2 is
4 minus 2, and y
is equal to 2.
When you put negative 2 here, y
is also equal to 2, so that
makes sense.
There's going to be some other
point way out here where they
also intersect.
There's also going to be some
other point way out here if we
keep making this parabola.
When y is equal to positive 4,
and you have negative 16 plus
6, you get negative 10.
So, positive 1, 2, 3, 4, and
then you go down 10.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
That looks like that might
be our other point of
intersection, so let me connect
this right there.
Our other point of intersection
looks to be right there.
If we just follow this red
line it looks like we
intersect there.
Let's verify that
it works out.
So 4, negative 10.
We know that that's on this
blue line, so let's see if
it's on this other line.
So negative 2 times 4 minus 2,
that is negative 8 minus 2,
which is equal to negative 10.
The point 4, negative 10,
is on both of them.
When x is equal to 4, y is
negative 10 for both equations
here, so they both definitely
work out.