If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:6:12

Quadratic systems: a line and a parabola

Video transcript

we're told the parabola given by y is equal to 3x squared minus 6x plus 1 and the line given by y minus X plus 1 equals 0 our graph so you can see the parabola here in red and we can see the line here in blue and the first thing they asked us is one intersection point is clearly identifiable from the graph what is it and 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 equals two and y equals one it seems to be the point two comma one so it's 2 comma 1 there and 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 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 gonna we're gonna have to try to solve this system of equations and this is interesting because this is a system of equations where one of the equations is not linear where it is a quadratic so let's see how we could go about doing that so let me write down the equations I have Y is equal to 3x squared minus 6x plus 1 and our next equation right over here Y y minus X plus 1 is equal to 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 as 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 1 and then hopefully I can solve for X so let's solve for y here it 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 one from both sides that was hard to see so and subtract one from both sides and so we are going to get Y and then all the rest of stuff cancels out is equal to X minus one and so now we can substitute X minus one back in for Y and so we get X minus one is equal to 3x squared minus 6x plus one now we can we want to get a zero 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 one to both sides and then what do we get on the left hand side we just get zero and on the right hand side we get 3x squared minus 7x plus 2 so this is equal to zero 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 & 2 3 times 2 and if this looks unfamiliar you can review factoring by grouping and can I think of those same two a plus B where it's going to be equal to negative 7 and actually negative 6 and negative 1 work so I can what I can do is I can rewrite this whole thing as 0 is equal to 3x squared and then instead of negative 7x I can write negative 6x and minus X and then I have my plus 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 is equal to so if I group these first two I can factor out a 3x so I'm going to get 3x times X minus 2 and in these second two I can factor out in these second two I can factor out a negative one so I have negative 1 times X minus 2 and then I can factor out a negative 2 I'll scroll down a little bit so I have some space so I have 0 is equal to if i factor out an X minus 2 I'm going to get X minus 2 times 3x minus 1 and so solution would be a situation where either of these is equal to 0 or I'll scroll down a little bit more so X minus 2 could be equal to 0 or 3x minus 1 is equal to 0 the point at which X minus 2 equals 0 is when X is equal to 2 and for 3 X minus 1 equals 0 add 1 to both sides you get 3x is equal to 1 or X is equal to 1/3 so we figured out the we already saw the solution where X is equals equals 2 that's this point right over here we already type that in but now we figured out the x value of the other solution so this is X is equal to 1/3 right over here so our x value is 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 could figure out what is X when or what is y when x is equal to 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 is equal to 1/3 minus 1 I'm just substituting that 1/3 back into this and so you get Y is equal to negative 2/3 and it looks like that as well Y is equal to negative 2/3 right over there so this is the point 1 3 comma negative 2/3 and we're done