WEBVTT

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If the discriminant is negative, then the zeros are going to be complex zeros,

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they're going to be imaginary zeros.

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So speaking of that, let this be any quadratic function. How many zeros are we

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guaranteed to have?

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Two. What kind? Two complex zeros. They might be real or they might be imaginary,

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but we definitely have two complex zeros. Agreed?

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What are the two zeros? To find zeros, what do we do? If we plug in x = 0, we get

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f(0) = a times 0 squared plus b times 0…

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That gives the y-intercept. We want zeros, so plug 0 into f(x), or y. Then how do we

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solve?

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By factoring, completing the square, or the quadratic formula. This is how we

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derived the formula: the two zeros are [−b ± √(b² − 4ac)]/(2a).

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Does that make sense? It is abstract because these are variables, but they are all

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numbers. These are the two zeros. I could write it this way.

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We agree? Okay.

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Let me put an absolute value inside the square root to make this correct.

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If b² − 4ac is negative, these zeros are imaginary.

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We get a square root of a negative number: the square root of its magnitude times

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√(−1). And √(−1) is i.

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The absolute value makes that negative discriminant positive inside the radical; the

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i comes from √(−1).

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These are the two zeros: −b/(2a) plus the square-root term. I will keep the absolute

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value there.

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And −b/(2a) minus the square-root term over 2a.

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These are two complex numbers.

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What's the only difference between these two?

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Plus versus minus. What do we call that? Conjugates.

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This is an important fact: if b² − 4ac is less than zero—negative—

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the two zeros are complex conjugates of one another.

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2i is a known zero of f(x).

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The leading coefficient is −3. Find the standard form of f(x).

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We know one complex zero. It's imaginary. And we know the leading coefficient.

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We know one complex zero. It's imaginary.

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Okay.

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So what does that tell us?

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Its complex conjugate is also a zero. The complex conjugate of 2i is also a zero.

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And what is the complex conjugate of two i?

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0 + 2i. Its conjugate is 0 − 2i, so −2i is also a zero.

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And now we're going to write it in what form? Factored form. When we know the zeros,

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we can write factored form.

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So it's going to be

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f(x) is the leading coefficient, −3, times (x − 2i)(x − (−2i)), which is (x − 2i)(x

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+ 2i).

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Now we just check. Okay. So that is the function, but it's written in factor form.

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And how are we going to get that into standard form?

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Distribute.

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Minus 4i².

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Those two terms add to zero, and i² is −1.

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So the product is x² + 4.

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And that's not quite in standard form.

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f(x) = −3x² − 12.

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And there it is.

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I could ask other questions: find the vertex, graph it, or what else?

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The axis of symmetry. That one is easy.

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I could have done no work. What's the axis of symmetry at the very beginning? Zero.

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Zero, because it's this number over two a.

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Let's try another one slightly more complicated. What was that first one I did? What

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was it?

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3 + 2i is a known zero.

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And

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the

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vertex.

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Let’s use the axis of symmetry.

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What do we know is also a zero? 3 − 2i is also a zero.

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So we can write factored form. What is the question asking? Good question.

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Write the standard form

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of f(x).

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We know the zeros, so we can almost write the factored form of f(x).

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We don’t have a leading coefficient, so leave an a there for now.

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It is x − (3 + 2i)

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times x − (3 − 2i).

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Let's multiply this out.

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We will figure out a. For now: a times (x − 3 − 2i) times (x − 3 + 2i).

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Then multiply it all out. There is a lot of distributing. How many terms? Nine.

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You could do it. X times X is X squared. X squared.

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−3x + 2ix.

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Minus three X.

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−3x + 9 −6i.

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−2ix + 6i

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−4i².

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Everything with i cancels, or i² turns into −1.

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2ix and −2ix cancel. If they didn’t, we multiplied something incorrectly. We have a

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times x²…

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−3x − 3x is −6x,

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and 9 + 4

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is 13.

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I don't think I gave you enough information.

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I don't think so. I messed up.

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There are actually many possibilities.

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This is ax² − 6ax + 13a. We agree.

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So we know.

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Negative 6 A.

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Over.

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Two a. This is our negative b— Sorry negative.

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This is equal to 3.

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No I didn't. I take that back. I screwed up.

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We don’t have enough information to answer the question.

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Okay. Yeah.

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Let’s instead say this is the leading coefficient.

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Then we substitute 3 here.

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The function would be 3x² − 18x + 39. Okay.

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If you are given just one real zero, you cannot determine the other zero.

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But given a nonreal complex zero, its conjugate is also a zero,

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so we know the other zero.

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A quadratic function has vertex (−3, 2) and passes through

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(−1, 14).

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Find the standard form.

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How we can do this.

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Okay. They give us the vertex.

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We can almost write vertex form; the only missing quantity is the leading

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coefficient. So we got to leave this as an A.

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x − (−3) is x + 3, squared, plus 2.

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And last thing I do is find the leading coefficient. Multiply that out.

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f(−1) = 14. That is what this ordered pair means. We plug that in here.

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14 equals a

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times (−1 + 3)², plus 2. And now we got an equation with just one variable.

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And we do a little bit of work.

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14 − 2

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equals a times 2 squared.

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That is 4a.

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Divide by 4.

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So this is the.

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This is the equation in what form.

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Vertex form. How do we get standard form? Just start multiplying out.

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I will go quickly: x² + 3x + 3x + 9, then plus 2.

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Plus 27 plus 2 is 29. Yes.

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OK. So if you're given some information.

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Suppose I gave you two real zeros

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and one ordered pair.

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Substituting that point lets us find the leading coefficient.

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OK.

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Going back last year. Where does.

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Where do y = 2x + 3 and y = −3x − 7 intersect?

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You might have done problems like this last year. Yes. The others like this.

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And we need to find.

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Where they intersect. We won’t finish this example, but we will go a little further.

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How will we go about doing this.

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OK.

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Yeah we're not going to try to graph them because it might not need a very nice

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point.

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We could if they met at a nice point and figured out. We have y = 2x + 3 and y = −3x

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− 7.

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We want an ordered pair that makes both equations true.

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And we have another name for that. What do we call it.

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A system of equations.

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OK. And last year there were like three ways to solve systems equations.

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Elimination, substitution, and…

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I think it was the only way.

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Yeah. Yeah. Which we don't do. OK. So for this what we do.

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Substitute. We take this guy. Substitute in right. This is y, so substitute it for

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this y. OK.

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We get 2x + 3 = −3x + 7. [Correction on screen: the original constant is −7.] And we

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solve this and get some number for X right.

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So is that our solution. No. No, that is only the x-coordinate. Substitute x into

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either original equation to find y.

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So we're going to do something similar but now with quadratics.

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We want to find where.

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They intersect.

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First off. How many possible solutions. How many possible.

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Times can intersect.

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Does this one open up or down.

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So it opens up. This one open up or open down down.

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With one opening up and one opening down, how many intersection points could there

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be? Two. Can we have just one.

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They can meet at a single point. Well.

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They could be at the vertex.

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OK. And could there be no solutions or no intersections. Yes. Yeah. If this guy was

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down here.

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So how are we going to do this.

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So. OK. Same thing as. Right. y = x² + 2x − 2 and y = −x²

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+ 4x + 10.

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And this is a system we're going to solve.

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OK.

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Systems are we are looking for solutions that are common to both of them. So the

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solution is going to be ordered pairs.

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And we're solve this by.

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Substitution. We're going to substitute this. Which is equal to Y into this Y or you

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could do it vice versa.

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x² + 2x − 2 = −x²

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+ 4x + 10.

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Oh now we have an equation in only one variable. And now we're going to solve this.

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It's just a quadratic equation. We get everything on one side.

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Add x², subtract 4x, and subtract 10.

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2x² − 2x − 12 = 0.

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And then you could solve by factoring completing the square quadratic formula. These

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have a common factor. Which should we do first?

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The two. Yeah get rid of that two. It will make it easier.

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And does this factor easily?

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It does. What's the factor?

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x − 3 times x + 2. The solutions are 3 and −2.

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Is that my answer? 3, −2? Those are X values.

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So to find the ordered pairs we plug them into. Either one of them. Let's plug it in

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here. What was the first one?

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f(3). 3² + 2 times 3 − 2.

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9 + 6 − 2

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is 13.

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And g of… What was the other one? Negative one? Negative two. Negative two.

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Is that negative two? No. So they intersect.

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(3, 13)

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and (−2,

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−2).

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Okay.

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We could have a quadratic and a linear. Do the same thing. Substitute the linear

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expression into the quadratic, or vice versa.

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And we solve that quadratic. If you end up with the irrational solutions.

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Substitution into the functions may look less tidy.

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Find the equation of the line

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That.

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Passes.

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through

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Both.

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vertices.

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The quadratic has a vertex, and the absolute-value function has a vertex.

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What is the Vertex of this one? This one? easy.

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(3, 1). This is in vertex form. Take the opposite of this number, paired with that

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number. How do we find the vertex of this one?

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−2. −2. And now we plug it in to find the y value.

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4 − 8 + 1. Is that −3? Yes.

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Okay, so the vertex here is

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(−2, −3). And then what did I ask for?

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The equation of the line through both vertices.

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I don't want to graph it. I want to find the equation.

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Okay, we need to find the slope first.

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Which is going to be right. (y₂ − y₁)/(x₂ − x₁).

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1 minus negative 3, over 3 minus negative 2. That is 4/5.

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Now we need the y value. We can then find the y intercept or

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The better way is point-slope form. Oh, yeah.

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Which point should I use? This point or this point?

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I think that was the point. It's nice.

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It doesn't matter.

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y − 1 = (4/5) times ( x

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minus 3).

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And that's it.

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Do I need to put it in slope-intercept form? No, it is already the equation of the

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line. No matter what form it's in.

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So we can answer all kinds of crazy questions now.