Graphing quadratics in factored form (video) | Khan Academy (2024)

Video transcript

- [Instructor] We're askedto graph the equation y is equal to one-half times x minus six times x plus two. So like always, pause this video and take out some graph paper or even try to do it ona regular piece of paper and see if you can graph this equation. All right now, let's workthrough this together. There's many different ways that you could attempt to graph it. May be the most basic istry out a bunch of x values and a bunch of y values andtry to connect the curve that connects all of those dots. But let's try to see ifwe can get the essence of this graph withoutdoing that much work. The key realization here withouteven having to do the math is if I multiply this out, if I multiplied x minussix times x plus two, I'm going to get a quadratic. I'm going to get x-squared, plus something, plus something else. And so this whole thingis going to be a parabola. We are graphing a quadratic equation. Now a parabola you might remember can intersect the x-axis multiple times. So let's see if we can find out where this intersects the x-axis. And the form that it's in, it's in factored form already, it makes it pretty straightforward for us to recognizewhen does y equal zero? Which are going to be thetimes that we're intersecting the x-axis. And then from that, we'llactually be able to find the coordinates of the vertex and we're going to be ableto get the general shape of this curve which isgoing to be a parabola. So let's think about it. When does y equal zero? Well to solve that we justhave to figure out when, if we want to know when y equals zero, then we have to solve for when does this expression equal zero? So let's just solve the equation. One half times x minus six, times x plus two is equal to zero. Now in previous videos we'vetalked about this idea. If I have the product of multiple things and it needs to be equal to zero, the only way that's going to happen is if one or more of these things are going to be equal to zero. Well one half is one-half, it's not going to be equal to zero. But x minus six could be equal to zero. So if x minus six is equal to zero, then that would make this equation true. Or if x plus two is equal to zero, that would also make this equation true. So the x values thatsatisfy either of these would make y equal zeroand those would be places where our curve isintersecting the x-axis. So what x value makesx minus six equal zero? Well you could add six to both sides, you're probably able todo that in your head, and you get x is equal to six. Or you subtract two from both sides here and you get x is equal to, these cancel out, you get x is equal to negative two. These are the two x values where y will be equal to zero. You can substitute it backinto our original equation. If x is equal to six, then this right over here isgoing to be equal to zero, and then y is going to be equal to zero. If x is equal to negative two, then this right over here isgoing to be equal to zero, and y would be equal to zero. So we know that our parabolais going to intersect the x-axis at x equalsnegative two right over there, and x is equal to six. These are our x-intercepts. So given this, how do wefigure out the vertex? Well the key idea here is to recognize that your axis ofsymmetry for your parabola is going to sit right betweenyour two x-intercepts. So what is the midpoint between, or what is the averageof six and negative two? Well, you could do that in your head. Six plus negative two isfour divided by two is two. Let me do that. So I'm just trying to find themidpoint between the point, let's use a new color. So I'm trying to find the midpoint between the point negative two comma zero and six comma zero. Well the midpoint, those are just the average of the coordinates. The average of zero and zerois just going to be zero, it's going to sit on the x-axis. But then the midpointof negative two and six or the average negativetwo plus six over two. Well let's see, that's four over two, that's just going to be two, so two comma zero. And you see that there. You could have done thatwithout even doing the math. You say okay, if I want to go right in between the two, I want to be two away from each of them. And so just like that,I could draw an axis of symmetry for my parabola. So my vertex is going to siton that axis of symmetry. And so how do I know what the y value is? Well I can figure out, I can substitute backin my original equation, and say well what is y equalwhen x is equal to two? Because remember the vertexhas a coordinate x equals two. It's going to be two comma something. So let's go back, let's see what y equals. So y will equal to one-half times, we're going to see when x equals two, so two minus six, times two plus two. Let's see, this is negative four, this is positive four. Negative four times four is negative 16. So it's equal to one-halftimes negative 16, which is equal to negative eight. So our vertex when x is equal to two, y is equal to negative eight. And so our vertex is goingto be right over here two common negative eight. And now we can draw the general shape of our actual parabola. It's going to look something like, once again this is a hand-drawn sketch, so take it with a littlebit of a grain of salt, but it's going to looksomething like this. And it's going to be symmetricaround our axis of symmetry. That's why it's calledthe axis of symmetry. This art program I have,there's a symmetry tool, but I'll just use this and there you go. That's a pretty good sketchof what this parabola is or what this graph is goingto look like which it is, an upward-opening parabola.

As an expert in mathematics, particularly in the field of algebra and graphing equations, I can confidently analyze the content of the provided video transcript. My understanding of the concepts involved allows me to elucidate the key points and provide additional insights.

The instructor in the video is guiding the audience through the process of graphing the quadratic equation (y = \frac{1}{2}(x - 6)(x + 2)). Here are the main concepts covered in the transcript:

  1. Graphing a Quadratic Equation: The given equation is a quadratic function in factored form, which is (y = \frac{1}{2}(x - 6)(x + 2)). The goal is to graph this quadratic equation on a coordinate plane.

  2. Realizing the Nature of the Graph: By multiplying the factored form, the instructor recognizes that the equation represents a quadratic function ((y = ax^2 + bx + c)), resulting in a parabola when graphed.

  3. Identifying x-Intercepts: The instructor explains that finding the x-intercepts (where (y = 0)) can help determine key points on the graph. By setting the equation equal to zero and solving for (x), they find the x-values where the parabola intersects the x-axis.

  4. Determining the Axis of Symmetry: The midpoint between the two x-intercepts gives the axis of symmetry for the parabola. The instructor calculates the average of the x-values to find the x-coordinate of the vertex.

  5. Finding the Vertex: Substituting the x-coordinate of the vertex back into the original equation, the instructor determines the y-coordinate of the vertex. The vertex is a critical point on the graph that lies on the axis of symmetry.

  6. Sketching the Parabola: Using the x-intercepts and the vertex, the instructor provides a hand-drawn sketch of the parabola, emphasizing its upward-opening nature. The symmetry of the parabola is highlighted around the axis of symmetry.

Overall, the instructor demonstrates a systematic approach to graphing a quadratic equation, combining algebraic manipulation and geometric understanding to visualize the parabola's key features.

Graphing quadratics in factored form (video) | Khan Academy (2024)

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