Quadratic systems: a line and a parabola | Equations | Algebra 2 | Khan Academy

We're told the parabola given by ( y = 3x^2 - 6x + 1 ) and the line given by ( y - x + 1 = 0 ) are graphed.

So you can see the parabola here in red and we can see the line here in blue. The first thing they ask us is, one intersection point is clearly identifiable from the graph. What is it? They want us to put it in here.

This is actually a screenshot from the exercise on Khan Academy, but I'm just going to write on it. If you're doing it on Khan Academy, you would type it in, but pause this video and see if you can answer this first part.

All right, so one intersection point is clearly identifiable from the graph. I see two intersection points. I see that one and I see that one there. This second one seems clearly identifiable because when I look at the grid, it looks clearly to be at a value of ( x = 2 ) and ( y = 1 ). It seems to be the point ( (2, 1) ).

So it's ( (2, 1) ) there. What's interesting about these intersection points is, because it's a point that sits on the graph of both of these curves, that means that it satisfies both of these equations, that it's a solution to both of these equations.

So the other one is to find the other intersection point. Your answer must be exact. So they want us to figure out this intersection point right over here. Well, to do that, we're going to have to try to solve this system of equations.

This is interesting because this is a system of equations where one of the equations is not linear; it is a quadratic. So let's see how we could go about doing that.

Let me write down the equations. I have ( y = 3x^2 - 6x + 1 ) and our next equation right over here, ( y - x + 1 = 0 ).

Well, one way to tackle, and this is one way to tackle any system of equations, is through substitution. So if I can rewrite this linear equation in terms of ( y ), if I can solve for ( y ), then I can substitute what ( y ) equals back into my first equation, into my quadratic one, and then hopefully I can solve for ( x ).

Let's solve for ( y ) here. Actually, let me color code it because this one is in red and this one is the line in that blue color. So let's just solve for ( y ). The easiest way to solve for ( y ) is to add ( x ) to both sides and subtract ( 1 ) from both sides.

That was hard to see, so we subtract ( 1 ) from both sides, and so we are going to get ( y ) and then all the rest of the stuff cancels out, is equal to ( x - 1 ).

Now we can substitute ( x - 1 ) back in for ( y ), and so we get ( x - 1 = 3x^2 - 6x + 1 ). Now we want to get a ( 0 ) on one side of this equation, so let's subtract ( x ). I'll do this in a neutral color now.

Let's subtract ( x ) from both sides and let's add ( 1 ) to both sides. Then what do we get? On the left-hand side, we just get ( 0 ), and on the right-hand side, we get ( 3x^2 - 7x + 2 ). So this is equal to ( 0 ).

Now we could try to factor this. Let's see, is there an obvious way to factor it? Can I think of two numbers ( a \times b ) that's equal to the product of ( 3 ) and ( 2 )? Three times two.

If this looks unfamiliar, you can review factoring by grouping. Can I think of those same two ( a + b ) where it's going to be equal to ( -7 )? Actually, ( -6 ) and ( -1 ) work.

So what I can do is I can rewrite this whole thing as ( 0 = 3x^2 ) and then instead of ( -7x ), I can write ( -6x - x ), and then I have my ( +2 ).

I'm just factoring by grouping. For those of you who are not familiar with this technique, you could also use a quadratic formula. So then ( 0 = 3x ) times ( x - 2 ).

In these second two, I can factor out ( -1 ), so I have ( -1 ) times ( x - 2 ).

Then I can factor out a ( -2 ). I'll scroll down a little bit so I have some space.

So I have ( 0 = ) if I factor out ( (x - 2) ), I'm going to get ( (x - 2)(3x - 1) ). So a solution would be a situation where either of these is equal to zero.

Or I'll scroll down a little bit more. So ( x - 2 ) could be equal to ( 0 ) or ( 3x - 1 ) is equal to ( 0 ). The point where ( x - 2 = 0 ) is when ( x = 2 ).

And for ( 3x - 1 = 0 ), add ( 1 ) to both sides, you get ( 3x = 1 ) or ( x = \frac{1}{3} ).

So we figured out the— we already saw the solution where ( x = 2 ). That's this point right over here; we already typed that in. But now we figured out the ( x ) value of the other solution, so this is ( x = \frac{1}{3} ) right over here.

So our ( x ) value is ( \frac{1}{3} ), but we still have to figure out the ( y ) value. Well, the ( y ) value is going to be the corresponding ( y ) we get for that ( x ) in either equation.

And I like to focus on the simpler of the two equations so we can figure out what is ( y ) when ( x = \frac{1}{3} ).

Using this equation, we could have used the original one, but this is even simpler; it's already solved for ( y ). So ( y = \frac{1}{3} - 1 ).

I'm just substituting that ( \frac{1}{3} ) back into this, and so you get ( y = -\frac{2}{3} ). And it looks like that as well.

( y = -\frac{2}{3} ) right over there. So this is the point ( \left( \frac{1}{3}, -\frac{2}{3} \right) ) and we're done.