Algebra 2 Solving Quadratic Equations by Graphing

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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 ax² + 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 2x² + 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 x²/3 - 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 -x² - 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) = -x² - 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 solution0637

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 x² - 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) = x² - 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 3² - 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: x² + x + 1 = 0.1071

The corresponding function is x² + 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: x² - 8 = 0.1256

The corresponding function is f(x) = x² - 8.1260

Thinking about standard form, ax² + 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 0² - 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