solving systems of linear and quadratic equations

#Solving systems of linear and quadratic equations plus

Immediately pop out at us, because they asked us toĭo it graphically. It will look something likeĭo they intersect? One point of intersection does That line will look something like- It's hard for my hand toĭraw that, but let me try as best as I can. To move up 2 in the y direction, and it looks like weįound one of our points of intersection. Move 2 in the x direction, we're going to move downĤ in the y direction. We're going to go 2 in the y-direction, and if we Our y-intercept is negativeĢ, so 0, 1, 2. Over here: y is equal to negative 2x minus 2. Like, and obviously it keeps going down in that direction. Right here, and then let me connect this. That second part is hard toĭraw- let me do it from here.

That, and let me just do the second part. Something- I was doing well until that second part -like You have negative 3, negativeģ, and then you have 3, negative 3. Negative 3 squared is positiveĩ, you have a negative out front, it becomes negative 9

#Solving systems of linear and quadratic equations plus

Plus 3, it becomes negative 3, and negative 3 will alsoīecome a negative 3. It with 3, as well- if we put a 3 there, 3 squared is 9. There, so 2 comma 2, and then you have a negative 2 comma 2. Negative there, so it's negative 4 plus 6 is 2.

Square it, then you have positive 4, but you have a So when x is 2, what is y? You have 2 squared, which isĤ, but you have negative 2 squared, so it's negative 4 X is equal to- let me just draw a little table here. Graph a couple of other points, just to see The vertex of this parabola is when x is equal to 0,Īnd y is equal to 6. Thing can take on is when x is going to be equal to 0. When you multiply it by a negative, so it's going This whole term right here isĪlways going to be negative, or it's always going Its maximum point? Let's think about thatįor a second. Opening parabola? You see that it's a negativeĬoefficient in front of the x squared, so it's going to be aĭownward opening parabola. It is going to be upward opening, or downward Know it's a parabola? That's because it's a quadraticįunction: we have an x squared term, a secondĭegree term, here. Is this going to be an upward opening- one, how did I Let's start- let me find a nice dark color to Here is a great on line graphing tool where you can experiment and get to know the properties of quadratic equations: Īlgebraically. Typically, one of the first things we do is set x=0 and see what value the function produces. With more practice, these properties will become part of what you know. 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. 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. that means that we will have the case that the equation will be something like this: a negative number + 6. So now our goal is to figure out what value of x produces the maximum point.ĪLL values of -x² are negative. That means the graph will have a maximum point. So no matter what value x has -x² will produce a negative number.įrom 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.įrom 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). The function is a member of the family of quadratic equations since the highest degree is 2.Īll quadratic equations produce a parabola as a graph.

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. 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.