Algebra (all content)
Quadratic systems: graphical solution
Sal solves a system of a quadratic equation and a linear equation by graphing both equations and looking for their intersections, and then checks the solution algebraically. Created by Sal Khan and Monterey Institute for Technology and Education.
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- What is the difference between a linear or non-linear equation?(26 votes)
- Notice in the video that the equation where x was multiplied by a constant (-2), the graph was a straight line. The line was sloped, but it was straight. It was "linear". The equation that had x^2 made a curve, not a straight line.(36 votes)
- couldn't -x^2 be interpreted as -(x^2) or (-x^2)? It seems to me like it could be either one, and which one you should use would depend on how you arrived at the equation y = -x^2 + 6. If, for instance, you somehow came up with the variable -x, and then you realized that to get your answer you need to square it, this would be the case of (-x^2) which would result in a positive number. Or if you had your variable x, and you needed to square it and take the negative of it, you would be doing -(x^2). I'm sorry I can't think of any specific examples, but am I right in thinking that this can be interpreted in 2 ways, depending on the context?(6 votes)
- You are correct. To avoid confusion, people will do -x^2 when they mean -(x^2) and do (-x)^2 when they mean that.(4 votes)
- How did you get the porabla at3:02
I'm still confused..(2 votes)
- I think I understand your question. The line describing the parabola is curved, not straight, but with only a few points drawn in, it's hard to see why.
If you went through the equation and filled in "x" with all the points in between the points that he graphed, you would see a curve shape appear. When you only have two points it's easy to see why it looks like it should be straight. Sal just skipped over doing 2.5, and then 2.25, and so on. The more points you graph, the more it looks l(7 votes)
- Please explain the ^ symbol.(3 votes)
- ^ symbol means "raised to the power of"
x^0=1 →x raised to the power of 0
x^1=1*x →x raised to the power of 1
x^2=1*x*x →x raised to the power of 2
x^3=1*x*x*x →x raised to the power of 3
so on and so forth(6 votes)
- To find the points of intersection, you could also set the two equations equal to one another which gives you x=-2, +4 - plug each back into the easier of the equations and you get your related y-values.(4 votes)
- Yup! That can be done! Good Eye!
However . . . . .
The video is about finding solutions by graphing. Analytic (algebraic) methods, such as you have done, are coming up.(2 votes)
- If x^2 represents an upward parabola and -x^2 represents a downward parabola, what represents a sideways parabola?(2 votes)
- Andrew is correct. Just to add to his answer:
"sideways parabolas" are not functions and very seldom do you have to deal with any one of those.
This is becouse for any for every x there should be max 1 y value. So if you put in say 5 in the equation then you should not get two results. Two results would be the case if you have a sideways parabola.(3 votes)
- At0:47to1:30, I dont get how you can assume x=0 for the turning point(3 votes)
- What you call the turning point we call a maximum or a minimum point, which as you observed is where the graph changes direction or turns.
The equation under consideration is y=-x² + 6, so let's take a look at it and see if we can do some basic analysis to figure out what properties the graph might have.
The function is a member of the family of quadratic equations since the highest degree is 2.
All quadratic equations produce a parabola as a graph.
The first term is -x². So no matter what value x has -x² will produce a negative number.
From that, we know that as x gets bigger and bigger, -x² will produce and even bigger negative number, which means the graph goes further and further down into the negative area of the graph.
From that we now know that the graph is concave down (kind of like the letter n, whereas concave up us more like the letter u). That means the graph will have a maximum point. So now our goal is to figure out what value of x produces the maximum point.
ALL values of -x² are negative. that means that we will have the case that the equation will be something like this: a negative number + 6. That means the maximum point at most can be 6 and that only happens when x=0 :: y=-x² + 6 = -0² + 6 = 0 + 6 = 6. Can you see how ANY other value of x will produce a value less than 6? (no matter what x is if it is not equal to zero it will be a negative number which means you will be taking away the negative number from 6.
With more practice, these properties will become part of what you know.
Typically, one of the first things we do is set x=0 and see what value the function produces.
Keep practicing and you will get it.
Here is a great on line graphing tool where you can experiment and get to know the properties of quadratic equations: https://www.desmos.com/calculator(3 votes)
- my question is that is zero a positive or a negative number? and if it is both how can we write it(1 vote)
- Zero is neither positive nor negative. It is the only number that is neither -- all other (real) numbers are either positive or negative. There are no numbers that are both positive and negative.(6 votes)
- How would one solve a system of equations with trig functions in it? For the system: 2cos(x)+3sin(y) + (xy^3)-(3xy-1)=0 2cos(y)+9sin(x)+ 2(xy^2)-(xy-5) =0 or similar you can't use substitution. It is possible that my example has infinite solutions or no solutions , but I hope you get the idea.(2 votes)
- Your example has infinitely many solutions.
But, in general, it is possible to find solutions to these kinds of problems, provided that you sufficiently constrain the domain.
The answers themselves will typically involve exponents with complex number, trig functions, etc. You won't typically have just a single answer.
But, this is very advanced material, far too complicated for this level of study.(3 votes)
- how do you notice the difference between linear and non-linear equations?(1 vote)
- It is possible to write a linear equation in the form of
y = mx + b
Note that this only requires that it is POSSIBLE to write a linear equation in the above fashion, not that is is currently written in that fashion. Here are some examples of other ways a linear equation might be written:
(3y + 2)/5 = 4x
y/2 + x/3 = 7
2y - 5x + 8 = 0
y-2 = 3(x-6)
y = 7(x+4)
(y-5)/2 + (x-7)/3 = 0
It is not possible to write a nonlinear equation in the above fashion. Frequently (but not always), at this level of study, a nonlinear equation is written in a form where there is a term with
xwith an exponent other than
1OR there are factors where x is in some value of x multiplied by another value of x. Examples of nonlinear equations are
y = 1/x + 5
y = (x-7)(x+2)
y = 5x² + 3
y= x³ + 4x - x
y = sin x(5 votes)
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.