### Solving Quadratic Equations by Graphing

- A quadratic equation has 2 real roots if its graph has 2 x-intercepts, one real root if it has 1 x-intercept (in this case, the graph is tangent to the x axis), and no real roots if it has no x-intercepts.
- If a root is not an integer, estimate the root by stating the two consecutive integers it lies between.
- A real number is a zero of the quadratic function f(x) if and only if it is a root of the equation f(x) = 0.

### Solving Quadratic Equations by Graphing

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Quadratic Equations 0:16
- Standard Form
- Example: Quadratic Equation
- Solving by Graphing 1:41
- Roots (x-Intercepts)
- Example: Number of Solutions
- Estimating Solutions 9:23
- Example: Integer Solutions
- Example: Estimating
- Example 1: Solve by Graphing 10:52
- Example 2: Solve by Graphing 15:10
- Example 1: Solve by Graphing 17:50
- Example 1: Solve by Graphing 20:54

### Algebra 2

### Transcription: Solving Quadratic Equations by Graphing

*Welcome to Educator.com.*0000

*In today's lesson, we are going to talk about solving equations by graphing.*0002

*I am going to introduce multiple methods for solving quadratic equations--some algebraically,*0006

*and this first one is using graphing techniques that we have learned earlier.*0011

*OK, first to review: a quadratic equation has the standard form ax ^{2} + bx + c = 0, where a does not equal 0.*0016

*And the reason a cannot equal 0 is that, if it did, this first term would drop out.*0027

*These are also sometimes called second-degree equations.*0036

*And when we are working on graphing these, we need to make sure that the quadratic equation is in this standard form.*0040

*So, just to give you a couple of examples: a quadratic equation could be 2x ^{2} + 4x - 3 = 0.*0048

*And this is in standard form, because notice: it says that a, b, and c need to be integers.*0060

*So, I have it in this form; and a, b, and c are integers here; a is 2; b is 4; and c is -3; so it is in standard form--*0066

*whereas, if you had something like x ^{2}/3 - x + 2 = 0, this is not standard form, since here, a is 1/3; it is not an integer.*0081

*Now, the solutions of quadratic equations, which are often called the roots of the quadratic equations, are the x-intercepts of the graph of the corresponding function.*0100

*And a quadratic equation can have 0, 1, or 2 real solutions.*0114

*And these are some concepts that were covered in greater depth in Algebra I, so we are just doing a brief review of them here--*0120

*of the different number of solutions or roots that can occur.*0128

*Let's first look at an example for a quadratic equation -x ^{2} - 2x + 3 = 0.*0132

*If you are asked to solve that, you need to go about it by graphing the corresponding quadratic function, which would be f(x) = -x ^{2} - 2x + 3.*0143

*Now, just by looking at this, I see that a is less than 0; a is negative.*0156

*So, I know that I am going to have a parabola that opens downward; and I am going to go ahead and find the vertex.*0162

*And the x-value at the vertex is -v/2a; so x equals...well, here b is -2; so that is negative -2,*0168

*over 2...a is -1; b is -2; so this is going to give me 2 times -1.*0181

*OK, so this is going to give me x equals...that is a negative and a negative, is a positive; and then I have -2...which equals -1...*0189

*So, the vertex is at x = -1; so then, let's find what the y-value is.*0203

*That is f(-1) = -(-1) ^{2} - 2(-1) + 3, so this equals negative...-1 squared is 1; -2...that is just 2 times -1; let's do it like that; + 3.*0210

*So, working our way down, this is going to give me a -1, and then this is going to be -2 times -1 plus 2 plus 3.*0237

*So, this is going to be...3 and 2 is 5; minus 1--that is 4; so the vertex is going to be at (-1,4).*0264

*And I already said that this opens downward, so that is a maximum.*0277

*So, (-1,4) is the vertex, and that is a maximum.*0282

*To graph it, I am going to need some other points; so let's say I have x and y values.*0287

*When x is -2, if you work through this, you will find that y is 3; when x is 0, you can see that y is also 3.*0295

*When x is 1, if you figure it out, you will get 1; so that is -1, minus 2; so that is going to be -3; plus 3, is 0.*0306

*When x is 2, substituting in here will give you -5.*0320

*So, let's stop for a second and look at these: here, it says if I find the x-intercepts of the graph, those are the solutions.*0325

*The solutions of the quadratic equation are the x-intercepts of the graph.*0336

*Recall that, when y equals 0, at that point, you have the x-intercept.*0342

*So, here is an x-intercept at (1,0); so I actually found one of the solutions--so let's go ahead and graph this out.*0348

*x is -2; y is 3; when x is 0, y is 3; when x is 1, y is 0 (that is one of my solutions); and then, when x is 2, y is down here at -5.*0359

*I don't even need to go that far with this, though, because you will see that all I am trying to do is find the solution.*0380

*And I know that my axis of symmetry is going to be here at x = -1.*0386

*Because of that, because I have an axis of symmetry and I have one solution, I can actually reflect over here to find the other solution.*0392

*So, by using symmetry...I have my vertex here; I have my axis of symmetry; if I have a point here, 1, 2 over,*0403

*I am going to have another point here at -1, 2 over from the x-value of this axis of symmetry.*0420

*So, here I have (-3,0); here I have (1,0); and these are my two solutions.*0427

*So, solutions are -3 and 1; x equals -3 and 1.*0440

*Again, solving by graphing: you graph out the parabola; you find the x-intercepts; and those are the solutions for the quadratic equation.*0452

*Here, I actually had two solutions; however, there can be situations where there are no real solutions.*0461

*Let's think about when this can occur.*0468

*Let's say that I found a graph like this: this is my parabola.*0470

*Well, here is my minimum; this is never going to intersect with the x-axis, so there are no x-intercepts, so there are no real solutions.*0475

*OK, or if I had, say, this down here; it is never going to intersect with the x-axis.*0492

*So, I have a situation with two real solutions; I have a situation with no real solutions; how about 1?*0499

*Well, think about when that can occur: if the x-intercepts are the solutions, I could have a situation where the vertex is on the x-axis.*0506

*In that case, there will be one real solution; if the vertex is along the x-axis--the vertex intersects the x-axis--it is only going to intersect in one place.*0517

*So, I only have one real solution--one unique solution, or one real root.*0534

*So, two solutions is your typical situation that you find a lot; however, you may have no real solutions (it never intersects);*0540

*or you may have where you have intersection in only one point; you have only one x-intercept,*0549

*because the vertex is along that x-axis; so there is one real root.*0556

*Now, a limitation of graphing is that it is sometimes hard to get an exact solution.*0563

*If I graph, and I find that this is what my parabola looks like, and it intersects right here and right here, at x = -4 and x = -2,*0570

*those are integers; if I graph carefully, I can find the solution.*0585

*However, that doesn't always occur; and that is a limitation of using this system.*0588

*For example, let's say I make a graph, and I find this: my parabola has x-intercepts right here.*0593

*Now, using graphing, I am not going to be able to just see an exact value for x, like here.*0604

*In this case, the best I can do is to estimate solutions.*0612

*One of my solutions is that x is greater than 2, but less than 3; so I have here that x is greater than 2 and less than 3.*0616

*My other x-intercept, right here, is that x is greater than 4, but it is less than 5.*0630

*So, if it turns out that the solutions are not integers, the best you can do is to estimate the solution*0637

*by determining integers that surround the solution; here it was 2 and 3--here it was 4 and 5.*0644

*OK, solve by graphing: I always take a look at what I have and find the corresponding function, which is x ^{2} - 4x.*0652

*And I see that here, a equals 1; since a is positive (a is greater than 0), I know that my parabola is going to open upward.*0666

*And that way, I have a sense of what this is going to look like before I even start.*0674

*The vertex is going to have an x-value at -b/2a; and I do know that, since this opens upward, the vertex is going to be a minimum.*0679

*OK, a is 1; b is -4; so, substituting in, this is negative...b is -4...over 2 times 1; so, x equals...*0695

*a negative and a negative...that is a positive, so that is 4 over 2, equals 2.*0714

*So, my x-value for the vertex is 2; now, to find what the corresponding y-value is: f(2):*0726

*that is going to be 2 squared, minus 4 times 2, or 4 minus 8, which is -4.*0736

*The vertex is going to be at (2,-4): and I already know that this opens upward and that this is a minimum.*0747

*Now, I need to get some points, x and y.*0758

*Let's start with 0, because that is an easy one: when x is 0, y is 0, which is kind of nice, because I have already found a solution.*0764

*So, when y is 0, that gives you the x-intercept; so that is one solution, and I may have another solution.*0773

*Right now, I have one solution.*0788

*When x is...let's do -1: -1 squared would be 1; -4 times -1 is...this is going to give me 1, and then here I have -1 times 4,*0792

*which is going to give me -4, so this is going to be 1 plus 4, which is going to give me 5.*0809

*Now, when x is...let's see...3, that is going to give me 9 minus 12, so that is -3.*0818

*OK, how about 4? When x is 4, that is 16, and then that is going to give me minus 4 times 4; 16 -16 is going to give me 0; that is another solution.*0831

*So, I actually found both of my solutions; but even if I didn't, I could have just graphed, reflected across, and found that other solution.*0849

*So, let's go ahead and graph this out: (0,0); (-1,5) is up here; (3,-3); and (4,0).*0856

*I don't even really need to worry about that point; I am just looking for the solutions, and I have found them.*0869

*Those are the x-intercepts; and that is going to head up that way; but all I really wanted to find was these two x-intercepts right there.*0873

*Those are my solutions, so the solutions that I have are x = 4 and x = 0.*0889

*So again, solving by graphing requires that you graph in order to find the x-intercepts.*0899

*And if the x-intercepts are integers (like they are here), you can get an exact solution using this method.*0905

*OK, Example 2: Solve by graphing--therefore, we need to graph this corresponding function, f(x) = x ^{2} - 6x + 9.*0911

*Starting out, we are just saying, "OK, a is greater than 0; that means that this parabola is going to open upward, and my vertex is going to be a minimum."*0921

*So, the vertex's x-value is -b/2a; here, a is 1; b is -6, because this is in standard form.*0929

*So, finding the vertex: that is -(-6)/2(1) = 6/2 = 3.*0948

*OK, so I have my x-coordinate for the vertex at 3; let's go ahead and find the y-coordinate, f(3).*0959

*This equals 3 ^{2} - 6(3) + 9, equals 9 - 18 + 9, equals 0.*0968

*So, the vertex is at (3,0); now, noticing that (3,0) is an x-intercept, I already have one solution.*0979

*I know that this is a minimum; and if I do a little bit more thinking about this, I don't even have to go any farther with finding points and graphing.*0993

*This is my vertex, right?--and I already said that this opens upward; and this is the minimum value--y is not going to get any smaller than this.*1006

*So, if this opens upward, it is going to look like this; and the x-intercept (or intercepts) are my solution.*1014

*But since this is the vertex, and the vertex is on the x-axis, I only have one solution; it is not going to cross at two points.*1025

*So, I only have one real root, or one real solution; and that solution is x = 3.*1038

*So, recall that you can have two solutions if it were to intersect twice;*1046

*you may have no real solutions; and here is the situation where the vertex is on the x-axis, which means there is only one real solution.*1053

*In the next example, we are asked to solve by graphing: x ^{2} + x + 1 = 0.*1071

*The corresponding function is x ^{2} + x + 1.*1079

*Now, a equals 1, and b equals 1; since a is positive, this is going to open upward.*1084

*I am going to start out by finding my vertex, which is -b/2a.*1093

*So, that is -1/2(1); that is -1/2; OK, so I have my vertex, which is -1/2; so I want to find f(-1/2),*1101

*equals (-1/2) ^{2} - 1/2 + 1; that is going to give me -1/2 times -1/2, is going to give me positive 1/4, minus 1/2, plus 1.*1119

*That is going to give me 1/4, minus 1/2 is -1/4; plus 1: 1 minus 1/4 is 3/4.*1136

*OK, so my vertex is (-1/2,3/4).*1145

*That is approximately here; that is -1/2, and then 3/4 is about here.*1153

*Now, always, before you go and find points and waste time, stop and think about the shape,*1157

*and generally what type of solutions you are going to have, based on the shape of the parabola.*1163

*I know that this opens upward; therefore, my vertex is going to be a minimum value.*1167

*So, this is the smallest y is going to get; so this is going to curve up like this.*1175

*And what I am looking for are my x-intercepts, because those are my solutions, or solution.*1182

*But here, I can see: this is never going to intersect with the x-axis.*1195

*So, no matter what values I find for x, y is never going to become 0.*1198

*So, an x-intercept will occur when y is 0; but the minimum value that I have for y is 3/4.*1204

*The range is that y is going to be greater than or equal to 3/4.*1211

*So, since y is never going to become 0, this will never intersect with the x-axis; there are no real solutions.*1215

*In the last example, we found that there was one real solution, because the x-intercept was the vertex.*1223

*And again, I didn't have to waste time graphing out the whole parabola; the same here.*1232

*So, before you find a bunch of x and y values and get an exact graph, stop and think:*1236

*do I have one real solution, no real solutions, or two real solutions?*1241

*And if you have one real solution, all you have to do is find the vertex.*1245

*If you have no real solutions, you are done; if you have two real solutions, then, of course, you need to finish out your graphing.*1248

*Again, solve by graphing: x ^{2} - 8 = 0.*1256

*The corresponding function is f(x) = x ^{2} - 8.*1260

*Thinking about standard form, ax ^{2} + bx + c, here my a-value is 1; there is no bx term,*1269

*and what that tells me is that b must be 0; so it dropped out.*1281

*Now, since a is 1, a is greater than 0; this is going to open upward, and my vertex is going to have an x-value at -b/2a.*1286

*Here, since b is 0, it really didn't matter what a is, because 0 divided by anything is 0.*1302

*So, x equals 0 at the vertex; in order to find what my y-value is, I am going to find f(0), which is 0 ^{2} - 8, or -8.*1312

*So, the vertex is at (0,-8); -2, -4, -6, -8; and it is at (0,-8), so it is right here.*1326

*And since this opens upward, it is a minimum; so I know that this is going to go up.*1339

*And since it is curving upward, I am going to have two x-intercepts and two real solutions.*1343

*If this opened downward, I would be done; I would stop right now.*1351

*But that is why I always think, "Does it open upward or downward?"*1355

*So, I do have two real solutions; what I need to do is get some x- and y-values and figure out the rest of this graph.*1357

*When x is 1, we are going to get 1 - 8; then y is -7; when x is 2, that is 4 - 8; that is -4.*1365

*When x is 3, that is 3 times 3 is 9, minus 8 is 1; we already have x is 0,*1375

*so I am not even going to bother doing that one, because we already have that point.*1385

*I can put it here, though: (0,-8), and that is the vertex.*1391

*Let's get some negative values: when x is -1, this becomes one, minus 7--that is -7.*1395

*When x is -2, -2 squared is 4, minus 8 is -4; and one more--when x is -3, -3 squared is 9, minus 8--that is 1; I have plenty of points.*1403

*2, 4, 6, 8...OK, so when x is 1, y is right about here at -7; when x is 2, y is -4; when x is 3, y is 1.*1417

*And then, I have my vertex already right here; this is the vertex.*1441

*When x is -1, y is right here: -7; and then, you could also just do this by reflection, too--this other half.*1446

*When x is -2, y is -4; when x is -3, y is 1.*1454

*OK, now I am going to stop here and look at what I have going.*1461

*And what I see is that I have the x-intercepts, because I have switched from y being negative to y being positive.*1467

*So, right here, I crossed over that x-axis.*1475

*However, these are not integers; and since I am using graphing as a technique to solve, if my solutions are not integers, the best I can do is estimation.*1484

*So, I am going to estimate; and I am going to see the two integers that surround this solution.*1495

*And this is actually between -4 and -3; so, the best I can do is say that x is greater than -3 and less than -2.*1502

*And so, you can see where that crossover point is; and that is the best you can do.*1518

*That is that somewhere between x being (and if it is not exact in the graph, you can always look right here) -2 and -3, y crossed over to become positive.*1528

*So, even if it is not exact what the numbers are right in here, you can look here.*1542

*At one point, I had that x was -2 and y was a negative value.*1547

*And then, at another point, when x was -3, y became a positive value.*1554

*So, at some value of x in here, y was 0.*1558

*The same on this side: I know that, when x was 2, y was a negative value.*1567

*I know that, when 3 was 3, y was a positive value; so somewhere between 2 and 3, y was 0, which is my x-intercept, which is my solution.*1573

*So, the solution is somewhere in between here; and then there is another solution somewhere in between here.*1583

*My other solution would be estimated as "x is greater than 2, but less than 3."*1591

*At the crossover point where y goes from being negative to being positive, somewhere in here, is the x-intercept.*1597

*So again, I couldn't get an exact solution because of graphing; however, I can estimate, based on surrounding integers.*1605

*So today on Educator.com, we covered a method for solving quadratic equations; and that is solving by graphing.*1613

*See you next time to cover more methods for solving such equations.*1619

2 answers

Last reply by: Jose Gonzalez-Gigato

Fri Jan 27, 2012 10:11 AM

Post by Jose Gonzalez-Gigato on November 30, 2011

While Example III has indeed no real number solutions, I believe the correct vertex is (- 1/2, 1 3/4).