]> Chapter 05 Notes

5: Quadratic Equations and Functions

5-1: Modeling Data with Quadratic Functions

A Quadratic Function is a function that can be written in the form $a x^{2} + b x + c$ , where a, b and c are all real numbers, and $a \neq 0$ . Each of the pieces has a name: $a x^{2}$ is the leading term, or quadratic term; a is the leading coefficient; $b x$ is the linear term; c is the linear term, or constant term, or the constant.

The graph of a quadratic function is a parabola:

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A parabola has several features that you should be able to identify…

Direction: a parabola either opens up or down. This is determined by the sign of a. If $a > 0$ , then the parabola opens up. If $a < 0$ , then the parabola opens down.

Vertex: either the highest or lowest point on the graph. For now, you'll have to look at the graph to find this—later, we'll see how we can find it just from the equation.

Axis of Symmetry: the line of symmetry for the parabola. This vertical line passes through the vertex. If you already know the location of the vertex, then the axis of symmetry is the x-coordinate of the vertex.

If you are given some points, and asked to determine the equation of the parabola, plug the points into the standard form ( $a x^{2} + b x + c$ ) to obtain a system of equations—which you know how to solve!

Examples

[1.] Identify the vertex and axis of symmetry of the following parabola:

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The vertex appears to be at $(- 2, - 3)$ , and the line of symmetry is $x = - 2$ .

[2.] Find a quadratic model for the following points:

x	0	1	2	3
y	-2	0	4	10

So $f (0) = a \cdot 0^{2} + b \cdot 0 + c = - 2 ⇒ c = - 2$ ; this makes things easier!

$f (1) = a \cdot 1^{2} + b \cdot 1 - 2 = 0 ⇒ a + b = 2$

$f (2) = a \cdot 2^{2} + b \cdot 2 - 2 = 4 ⇒ 4a + 2b = 6$

That's all I need…I've now got two equations and two variables; I can solve this system!

${\begin{matrix} 2 a + 2 b = 4 \\ 4 a + 2 b = 6 \end{matrix} ⇒ 2 a = 2 ⇒ a = 1$ ; so $1 + b = 2 ⇒ b = 1$ !

The quadratic model is $y = x^{2} + x - 2$ .

5-2: Properties of Parabolas

From the standard form of a parabola ( $a x^{2} + b x + c$ ), we can determine several things:

The sign of a determines whether the parabola opens up ( $a > 0$ ; like the pictures above) or opens down ( $a < 0$ ).
The x-coordinate of the vertex (and the line of symmetry) is $- \frac{b}{2 a}$ .
The y-intercept is $(0, c)$ .

If the parabola opens up, then the y-coordinate of the vertex is the minimum value of the function. If the parabola opens down, then the y-coordinate of the vertex is the maximum value of the function.

The locations of the vertex and the y-intercept are often enough to get a good graph of a parabola…you can always plot more points, if necessary.

Examples

[3.] Graph the function $y = - x^{2} + 2 x - 3$ . Locate the vertex and axis of symmetry.

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The vertex is at $(1, - 2)$ , and the line of symmetry is $x = 1$ .

[4.] An object is launched into the air. Its height (in feet) after t seconds is given by $h (t) = - 16 t^{2} + 38.4 t + 0.96$ . How high up does this object go?

Let's look at the graph:

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A little zooming finds the maximum height to be 24 feet (at 1.2 seconds).

5-3: Translating Parabolas

Back in Chapter 2, we learned how to graph absolute value functions by using a template—by stretching and translating a parent function. It turns out that parabolas can also be graphed this way!

The template that most people write is $y = a {(x - h)}^{2} + k$ —called the vertex form of a parabola. The a does exactly what it did for absolute value functions—it multiplies the y-values; it stretches/shrinks/flips the graph. The h does exactly what my c did—moves the graph left and right. The k does exactly what my d did—moves the graph left and right. Note that the coordinates of the vertex are $(h, k)$ . Also, the a from standard form is the same as the a in vertex form.

I write this template as $y = a {(x - c)}^{2} + d$ .

At this point, you'll probably have to actually find the coordinates of the vertex in order to write the equation in vertex form.

Examples

[5.] Find the equation of the parabola below.

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The vertex is at $(- 2, - 6)$ , so I know most of the equation: $y = a {(x + 2)}^{2} - 6$ . To find the value of a, I need one more point…how about the y-intercept? It looks like $(0, 6)$ : $6 = a {(0 + 2)}^{2} - 6 ⇒ 6 = 4 a - 6 ⇒ 12 = 4 a ⇒ 3 = a$ . The equation is $y = 3 {(x + 2)}^{2} - 6$ .

[6.] Identify the vertex and line of symmetry of the parabola with equation $y = - 2 {(x + 5)}^{2} + 8$ .

Easy! The vertex is at $(- 5, 8)$ , which puts the line of symmetry at $x = - 5$ .

[7.] Write the equation $y = 3 x^{2} + 12 x - 1$ in vertex form.

Using the information from the previous section, the x-coordinate of the vertex is at $- \frac{12}{2 (3)} = - \frac{12}{6} = - 2$ . Plugging back into the equation, I find the y-coordinate to be $y = 3 {(- 2)}^{2} + 12 (- 2) - 1 = 3 \cdot 4 - 24 - 1 = - 13$ .

The equation must look like $y = a {(x + 2)}^{2} - 13$ . To find the value of a, I need another point—how about the y-intercept again? When $x = 0$ , $y = - 1$ (plug zero into the original equation), so that gives me $- 1 = a {(0 + 2)}^{2} - 13 ⇒ - 1 = 4 a - 13 ⇒ 12 = 4 a ⇒ 3 = a$ .

Finally, vertex form: $y = 3 {(x + 2)}^{2} - 13$ .

5-4: Factoring Quadratic Expressions

Factoring is important in mathematics. Together with some other properties of real numbers, factoring allows us to solve many problems without a calculator, or to see relationships that otherwise would remain hidden.

To get an idea about what we need to do to factor, let's assume that we've got a parabola already factored— $(m x + n) (p x + q)$ . FOIL this out and look at what you get: $(m p) x^{2} + (n p + m q) x + (n q)$ . That makes $a = m p$ , $b = n p + m q$ and $c = n q$ . The key to factoring is finding those two numbers— $n p$ and $m q$ —that multiply to give you ac ( $m n p q$ ) and add to give you b ( $n p + m q$ ).

This process of multiplying a times c, factoring that, and then choosing the factors that add to give you b, is called the Master Product Method. If you're having trouble with factoring, then it's okay to use this method explicitly; however, you will want to wean yourself off of it, so that you can factor quadratics without actually having to test a bunch of cases.

Examples

[8.] Factor $x^{2} - 7 x + 10$ .

There is nothing in common to start with, so on to the MPM.

I need numbers that multiply to give me 10, and add to give me -7. The factors of 10 are 1 and 10, or 2 and 5 (both positive or both negative). I notice that -2 plus -5 equals -7!

$x^{2} - 7 x + 10 = x^{2} - 2 x - 5 x + 10$

Now, look at those in pairs, and factor out any common items.

$x^{2} - 2 x - 5 x + 10 = (x^{2} - 2 x) - (5 x - 10) = x (x - 2) - 5 (x - 2)$

(note the negative trick in there!)

If it's going to work, you should see two factors that are the same…take them out, and write down what's left.

$x (x - 2) - 5 (x - 2) = (x - 2) (x - 5)$

All done!

[9.] Factor $12 x^{2} + 6 x + 18$ .

I note that all of the coefficients are multiples of 6!

$12 x^{2} + 6 x + 18 = 6 (2 x^{2} + x + 3)$

Now, I'll try to factor what's left: I need numbers that multiply to give me 6, and add to give me 1. The possible pairs are 1 and 6, or 2 and 3…but since both must be positive (or negative), neither pair add to give me 1. Therefore, the remainder can't be factored! The answer I gave above is all you can do.

[10.] Factor $2 x^{2} - x - 6$ .

Nothing in common…I need numbers that multiply to $(2) (- 6) = - 12$ and add to -1.

The choices are 1 and 12, 2 and 6, or 3 and 4. Note that within each pair, one number must be positive and the other must be negative.

3 and 4 must be the pair! Specifically, positive 3 and negative 4.

Now: split the middle term, group, then factor.

$2 x^{2} - x - 6 = 2 x^{2} - 4 x + 3 x - 6 = (2 x^{2} - 4 x) + (3 x - 6) = 2 x (x - 2) + 3 (x - 2)$

So $2 x^{2} - x - 6 = (x - 2) (2 x + 3)$ .

[11.] Factor $7 x^{2} - 28$ .

Take out the common value: $7 x^{2} - 28 = 7 (x^{2} - 4)$ .

Now I need two numbers that multiply to give me -4 and add to give me zero…the choices are 1 and 4, or 2 and 2. Remember that to multiply out to -4, one of the numbers must be positive, and the other must be negative.

A-ha! How about 2 and -2?

$7 (x^{2} - 4) = 7 (x^{2} - 2 x + 2 x - 4)$

Keep going…

$7 (x^{2} - 2 x + 2 x - 4) = 7 (x (x - 2) + 2 (x - 2)) = 7 (x - 2) (x + 2)$

5-5: Quadratic Equations

There are several methods for solving quadratic equations. First, and perhaps easiest, is graphing. I have no more to say about that method.

Second, if $b = 0$ , is to go ahead and isolate x, then take the square root of both sides of that expression. Just remember that when you take a square root, you must put a plus-or-minus symbol in front of the variable.

The most general technique is to use the zero-product property of the real numbers—if two numbers are multiplied to give zero, then one (or both) of those numbers must be zero. If $a b = 0$ , then either $a = 0$ or $b = 0$ .

To make use of this property, manipulate the equation so that one side is zero, then factor the quadratic. One of the factors must equal zero! Set each of these factors equal to zero and solve.

Examples

[12.] Solve graphically: $x^{2} - x = 1$ .

Graph $y = x^{2} - x$ and $y = 1$ :

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These graphs intersect at $x = - 0.618$ and $x = 1.618$ . These are the solutions to the equation.

[13.] Solve by taking the square root: $2 x^{2} = 8$ .

$2 x^{2} = 8 ⇒ x^{2} = 4 ⇒ x = \pm \sqrt{4} = \pm 2$ .

[14.] Solve by factoring: $5 x^{2} = 15 x$ .

To solve by factoring, you've got to get a zero on one side: $5 x^{2} = 15 x ⇒ 5 x^{2} - 15 x = 0$ .

Now, factor: $15 x ⇒ 5 x^{2} - 15 x = 0 ⇒ (5 x) (x - 3) = 0$ .

Using the zero product property, either $5 x = 0$ (which makes $x = 0$ ), or $x - 3 = 0$ , in which case $x = 3$ .

The solutions are $x = 0$ and $x = 3$ .

5-6: Complex Numbers

Before we discuss one (or two) final method(s) of solving quadratics, we need to investigate one last kind of numbers—complex numbers.

Quick—solve $x^{2} + 1 = 0$ .

Got it? No solution, you say? Well, you're correct, if you limit yourself to real numbers.

However, being able to give a name to the solutions of the previous equation turns out to be a really good thing…thus, we assign a name to those numbers. This name does not represent an actual measurement; it does not correspond to a number that occurs naturally in our universe. However, mathematicians do not feel constrained by reality!

(I can name at least three instances of mathematics that were considered "irrelevant" and/or "unnatural" until a use was later found for them!)

Let the symbol i be defined as the principal square root of negative one. The notation: $i \equiv \sqrt{- 1}$ . This symbol is called the imaginary number (or imaginary unit).

With this, we can solve $x^{2} + 1 = 0$ : $x^{2} + 1 = 0 ⇒ x^{2} = - 1 ⇒ \pm x = \sqrt{- 1} ⇒ x = \pm i$ .

The imaginary number can be added, multiplied, divided, subtracted, etc. with real numbers. This creates other numbers, called complex numbers.

A complex number is of the form $a + b i$ , where a and b are real numbers. Note that the textbook calls these imaginary numbers…and many people would call these imaginary numbers. There is some disagreement about what constitutes imaginary, and what is complex. I think I've made my viewpoint clear.

The real part of a complex number is a. The imaginary part of a complex number is b. Notation: if $z = a + b i$ , then $Re (z) = a$ and $Im (z) = b$ .

Complex numbers can be added, subtracted, multiplied, divided, etc. just like real numbers. For the most part, this works just like you'd expect. Dividing is harder; we'll not bother with that right now. Absolute value forces you to realize what absolute value really is…remember that it is not something that "makes numbers positive." Absolute value is a magnitude—distance to the origin. In order to measure the absolute value of a complex number, you must find its distance to the origin. To do that, you've got to be able to graph them.

Graphing a complex number requires a complex plane (or Argand plane; or Argand diagram). To do this, mark the horizontal axis as a (or real) and the vertical axis as b (or imaginary). Thus, the complex number $a + b i$ is plotted at $(a, b)$ .

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The distance to the origin is found by the Pythagorean Theorem!

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$d^{2} = a^{2} + b^{2} ⇒ d = \sqrt{a^{2} + b^{2}}$ . Thus, $| a + b i | = \sqrt{a^{2} + b^{2}}$ .

Note: the absolute value of a complex number is (should be) called the modulus.

Examples

[15.] Simplify: $\sqrt{- 16}$ .

$\sqrt{- 16} = \sqrt{(16) (- 1)} = \sqrt{16} \cdot \sqrt{- 1} = 4 \cdot i$ .

[16.] Write in standard form: $3 - \sqrt{- 25}$ .

$3 - \sqrt{- 25} = 3 - \sqrt{(25) (- 1)} = 3 - \sqrt{25} \cdot \sqrt{- 1} = 3 - 5 \cdot i$ .

[17.] Evaluate: $| 2 + 5 i |$

$| 2 + 5 i | = \sqrt{2^{2} + 5^{2}} = \sqrt{4 + 25} = \sqrt{29}$ .

[18.] Evaluate: $(5 + 2 i) + (6 - 3 i)$ .

$(5 + 2 i) + (6 - 3 i) = (5 + 6) + (2 - 3) i = 11 - i$ .

[19.] Evaluate: $(2 - 3 i) (4 - i)$ .

FOIL! $(2 - 3 i) (4 - i) = (2 \cdot 4) + (2 i) + (- 3 i \cdot 4) + (- 3 i \cdot i)$ .

Now, this needs some work. $(2 \cdot 4) + (2 i) + (- 3 i \cdot 4) + (- 3 i \cdot i) = 8 + 2 i - 12 i - 3 i^{2}$ .

One last trick: since $i = \sqrt{- 1}$ , $i^{2} = - 1$ . So $8 + 2 i - 12 i - 3 i^{2} = 8 - 10 i + 3 = 11 - 10 i$ .

[20.] Solve: $2 x^{2} + 1 = - 17$ .

$2 x^{2} + 1 = - 17 ⇒ 2 x^{2} = - 18 ⇒ x^{2} = - 9 ⇒ x = \pm \sqrt{- 9} = \pm 3 i$ .

5-7: Completing the Square

We've discussed how to factor a quadratic equation when $b = 0$ —but what if it isn't? We need some more techniques.

First, let's look at what happens when you expand ${(x - y)}^{2}$ : ${(x - y)}^{2} = x^{2} - 2 x y + y^{2}$ . Notice that the middle term is double the product of the original two terms. This is important to notice for the method I am about to introduce.

The method is called completing the square. There are at least two variations of this; I'll just do one here (the variation for solving). First, move the constant term (c) to the other side of the equation—in other words, get it so that all of the terms with an x are on one side, and everything else is on the other side. Next, factor out the a (if there is one) from the x terms. Now, the real trick: take half of what had been b and square it. Add this number inside the parentheses on the left side; multiply this number by a and add it on the right side. The effect of this trick is twofold: first, since you've really added the same number to both sides (really!), you haven't changed the equation. Second, this trick puts the stuff in the parentheses on the left side into the form $x^{2} - 2 x y + y^{2}$ , which can be factored!

Okay, okay—that's pretty hard to follow generically. Read the examples that follow.

Also, note that if you move the constants back over to the side with x as you near the end, you'll have the vertex form of the parabola—you'll be able to tell the coordinates of the vertex just from the equation!

Examples

[21.] Complete the square: $x^{2} - 12 x + ?$

Take half of b and square it: ${(\frac{- 12}{2})}^{2} = {(- 6)}^{2} = 36$

That's it! $x^{2} - 12 x + 36$

[22.] Solve by completing the square: $2 x^{2} + 4 x - 7 = 9$ .

Factor the a from the first two terms: $2 (x^{2} + 2 x) - 7 = 9$ .

Take half of the new b and square it: ${(\frac{2}{2})}^{2} = 1^{2} = 1$ .

Add that inside the parentheses; multiply it by a before adding it to the other side: $2 (x^{2} + 2 x + 1) - 7 = 9 + 2$ .

Collapse: $2 {(x + 1)}^{2} - 7 = 11$ .

Solve: $2 {(x + 1)}^{2} = 18 ⇒ {(x + 1)}^{2} = 9 ⇒ x + 1 = \pm 3 ⇒ x = - 1 \pm 3$ .

The answers are $x = - 1 - 3 = - 4$ and $x = - 1 + 3 = 2$ .

[23.] Write the equation in vertex form by completing the square: $y = 4 x^{2} - 16 x + 17$ .

Factor: $y = 4 (x^{2} - 4 x) - 17$ .

Magic Number: ${(\frac{- 4}{2})}^{2} = {(- 2)}^{2} = 4$ .

Add. $y + 16 = 4 (x^{2} - 4 x + 4) + 17$ .

Collapse and solve for y: $y = 4 {(x - 2)}^{2} + 17 - 16 = 4 {(x - 2)}^{2} + 1$ .

5-8: The Quadratic Formula

Those of you who are sufficiently nerdy (like me) might wonder: if completing the square is really just an algorithm (hence my attempt at a generic description of how to do it), then why can't we just get to the answer and write that down as a formula?

The answer is: we can. And I will!

(hang on—this will get a little bumpy)

$a x^{2} + b x + c = 0$

$a (x^{2} + \frac{b}{a} x) = - c$

$a (x^{2} + \frac{b}{a} x + \frac{b^{2}}{4 a^{2}}) = - c + \frac{b^{2}}{4 a}$

$a {(x + \frac{b}{2 a})}^{2} = - \frac{4 a c}{4 a} + \frac{b^{2}}{4 a}$

${(x + \frac{b}{2 a})}^{2} = \frac{- 4 a c + b^{2}}{4 a^{2}}$

$x + \frac{b}{2 a} = \pm \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}}$

$x = - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{\sqrt{4 a^{2}}}$

$x = - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

The Formula

…and there you have it. The quadratic formula.

If $a x^{2} + b x + c = 0$ , then $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ .

Neato, eh?

The Discriminant

Now, there is just one tiny little detail…square roots can be a problem. Especially if you try to take the square root of a negative number…

Hence, the discriminant: $d = b^{2} - 4 a c$ . If $d > 0$ , then everything works out just fine, and we get two answers for x. If $d = 0$ , then the radical disappears, and we get just one number ( $- \frac{b}{2 a}$ ). However, if $d < 0$ , then we don't get any real solutions…we get complex (non-real) solutions!

Examples

[24.] Solve by the quadratic formula: $3 x^{2} + 2 x = 5$ .

First, satisfy the conditions: $3 x^{2} + 2 x = 5 ⇒ 3 x^{2} + 2 x - 5 = 0$ .

Now, go! $x = \frac{- 2 \pm \sqrt{2^{2} - 4 (3) (- 5)}}{2 (3)} = \frac{- 2 \pm \sqrt{4 + 60}}{6} = \frac{- 2 \pm \sqrt{64}}{6} = \frac{- 2 \pm 8}{6}$ .

The answers are $x = \frac{- 2 - 8}{6} = \frac{- 10}{6} = - \frac{5}{3}$ and $x = \frac{- 2 + 8}{6} = \frac{6}{6} = 1$ .

[25.] Evaluate the discriminant of $2 x^{2} - 5 x + 7 = 0$ . How many solutions (and of what type) does the equation have?

The discriminant is ${(- 5)}^{2} - 4 (2) (7) = 25 - 56 = - 31$ . Since this is negative, the equation has two non-real solutions.

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