elementary functions quadratic functions in the last ...kws006/precalculus/2.1...this is the...

Elementary FunctionsPart 2, Polynomials

Lecture 2.1a, Quadratic Functions

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 1 / 35

Quadratic functions

In the last lecture we studied polynomials of simple form f(x) = mx + b.Now we move on to a more interesting case, polynomials of degree 2, thequadratic polynomials.

Quadratic functions have form f(x) = ax2 + bx + c where a, b, c are realnumbers and we will assume a 6= 0.

The graph of a quadratic polynomial is a parabola.


Quadratic functions

The graph of a quadratic polynomial is a parabola.Here is the standard parabola y = x2.


Quadratic functions

All of the graphs of quadratic functions can be created by transforming theparabola

y = x2.

Just as we have standard forms for the equations for lines (point-slope,slope-intercept, symmetric), we also have a standard form for a quadraticfunction.

Every quadratic function can be put in standard form,

f(x) = a(x− h)2 + k

where a, h and k are constants (real numbers.)


Quadratic functions

The standard form for a quadratic polynomial is

f(x) = a(x− h)2 + k (1)

From our understanding of transformation, beginning with the simpley = x2,if we shift that parabola to the right h units,(replacing x by x− h)

stretch it vertically by a factor of a(multiplying on the outside of the parentheses by a)

and then shift the parabola up k units,(adding k on the outside of the parentheses)

we will have the graph for y = a(x− h)2 + k.


Quadratic functions

The turning point (0, 0) on the parabola y = x2 is called the vertex of theparabola;it is moved by these transformations to the new vertex (h, k).

If we are given the equation f(x) = ax2 + bx + c for a quadratic function,we can change the function into the above form by first factoring out theterm a so that

f(x) = ax2 + bx + c = a(x2 + bax + c

a).

We then complete the square on the expression x2 + bax + c

a

Let me explain completing the square....


Completing the square

A major tool for solving quadratic equations is to turn a quadratic into anexpression involving a sum of a square and a constant term.

This technique is called “completing the square.”

Recall that if we square the linear term x + A we get

(x + A)2 = x2 + 2Ax + A2.

Most of us are not surprised to see the x2 or A2 come up in the expressionbut the expression 2Ax, the “cross-term” is also a critical part of ouranswer.

In the expansion of (x + A)(x + A) we summed Ax twice.


Completing the square – the main idea

Here is a test of our understanding of the simple equation(x + A)2 = x2 + 2Ax + A2. Fill in the blanks:

1 (x + 2)2 = x2 + 4x + 4

2 (x + 5)2 = x2 + 10x + 25

3 (x + 1)2 = x2 + 2x + 1

4 (x + − 1)2 = x2−2x + 1

5 (x + − 3)2 = x2−6x + 9

6 (x +7

2)2 = x2 + 7x +

49

4



Applying this idea.If one understands this simple idea, that we can predict the square bylooking at the coefficient of x, then we can rewrite any quadraticpolynomial into an expression involving a perfect square.

For example, sincex2 + 4x + 4 = (x + 2)2

then a polynomial that begins x2 + 4x must involve (x + 2)2.

For example:

x2 + 4x = (x + 2)2 − 4

x2 + 4x + 7 = (x + 2)2 + 3

x2 + 4x− 5 = (x + 2)2 − 9



Since x2 − 6x is the beginning of

x2 − 6x + 9 = (x− 3)2

thenx2 − 6x = (x− 3)2 − 9

and sox2 − 6x + 2 = (x− 3)2 − 7,

x2 − 6x + 10 = (x− 3)2 + 1,

x2 − 6x− 3 = (x− 3)2 − 12,




x2 − 3x +9

4= (x− 3

2)2

then

x2 − 3x = (x− 3

2)2 − 9

4.



This is useful if we are trying to put an equation y = f(x) of a quadraticinto the standard form y = a(x− h)2 + k.

In the next presentation, we apply the concept of “completing the square”to put a quadratic into standard form and then we will use that form todevelop the quadratic formula.

(END)



Lecture 2.1b, Applications of completing the square

Dr. Ken W. Smith


2013



In the last lecture we discussed quadratic functions and introduced theconcept of “completing the square”, rewriting

x2 + 4x + 7 = (x + 2)2 + 3

x2 + 4x− 5 = (x + 2)2 − 9

x2 − 6x + 9 = (x− 3)2

x2 − 6x = (x− 3)2 − 9.

x2 − 6x + 2 = (x− 3)2 − 7,

x2 − 6x + 10 = (x− 3)2 + 1,

x2 − 6x− 3 = (x− 3)2 − 12.Smith (SHSU) Elementary Functions 2013 14 / 35



x2 − 3x +9

4= (x− 3

2)2

then

x2 − 3x = (x− 3

2)2 − 9

4.



For example, ify = x2 + 4x + 7

is the equation of a parabola then we can complete the square, writingx2 + 4x = (x + 2)2 − 4 and sox2 + 4x + 7 = (x + 2)2 − 4 + 7 = (x + 2)2 + 3. Our equation is now

y = (x + 2)2 + 3.

This is the standard form for the quadratic. We see that the vertex of theparabola is (−2, 3).

We can transform the graph of y = x2 into the graph of y = (x + 2)2 + 3by shifting the graph of y = x2 two units to the left and then shifting thegraph up 3 units.



An extra step occurs if the coefficient of x2 is not 1.

How do we complete the square on something like 2x2 + 8x + 15?Once again we focus on the first two terms, 2x2 + 8x.Factor out the coefficient of x2 so that

2x2 + 8x = 2(x2 + 4x).

Now complete the square inside the parenthesis so thatx2 + 4x = (x2 + 4x + 4)− 4 = (x + 2)2 − 4. Therefore

2x2 + 8x = 2(x2 + 4x) = 2[(x + 2)2 − 4] = 2(x + 2)2 − 8.

If y = 2x2 + 8x + 15 then completing the square gives

y = 2x2 + 8x + 15 = 2(x + 2)2 − 8 + 15 = 2(x + 2)2 + 7 .



What is the vertex of the parabola with equation y = 2x2 + 8x + 15?

Solution. Complete the square to write

y = 2x2 + 8x + 15 = 2(x + 2)2 + 7.

The parabola with equation y = 2(x + 2)2 + 7 has vertex (−2, 7).

To transform the parabola y = x2 to the graph of y = 2(x + 2)2 + 7

1 first shift y = x2 left by 2 units

2 then stretch the graph vertically by a factor of 2

3 and then slide the graph up 7 units.



Find the standard form for a parabola with vertex (2, 1) passing through(4, 5).

Solution. A parabola with vertex (2, 1) passing through (4, 5) hasstandard form y = a(x− h)2 + k where (h, k) is the coordinate of thevertex.Here h = 2 and k = 1 and so we know y = a(x− 2)2 + 1.

By plugging in the point (4, 5) we see that

5 = a(4− 2)2 + 1

and so a = 1.

So our answer isy = 1 (x− 2 )2 + 1



Find the vertex of the parabola with graph given by the equation

y = 2x2 − 12x + 4.

Solution. Completing the square, we see that

2x2 − 12x + 4 = 2(x2 − 6x) + 4 = 2((x− 3)2 − 9) + 4

= 2(x− 3)2 − 18 + 4 = 2(x− 3)2 − 14.

So the vertex is (3,−14).



Describe, precisely, the transformations necessary to move the graph ofy = x2 into the graph of y = 2x2 − 12x + 4, given on the previous slide.

Solution. Since the standard form for the parabola is y = 2(x− 3)2 − 14(from the previous slide) then we must do the following transformations, inthis order:

1 Shift y = x2 right by 3.

2 Stretch the graph by a factor of 2 in the vertical direction.

3 Shift the graph down by 14.

In the next presentation we discuss the quadratic formula(END)



Lecture 2.1c, The Quadratic Formula

Dr. Ken W. Smith


2013


The quadratic formula

We obtain a nice general formula for solutions to a quadratic equation ifwe complete the square on a general, arbitrary quadratic function. Let’ssee how this works.

Consider the general quadratic function

f(x) = ax2 + bx + c

where a, b and c are unknown constants.

We can put this function into standard form by completing the square.

First we factor out a from the first two terms,

f(x) = a(x2 +b

ax) + c.


The quadratic formula, continued

We complete the square on

f(x) = ax2 + bx + c

by factoring a from the first two terms,

f(x) = a(x2 +b

ax) + c.

Then we complete the square on x2 + bax. Since x2 + b

ax is the beginning

of the expression for (x + b2a)2 we compute

(x +b

2a)2 = x2 +

b

ax +

b2

4a2.

Subtracting b2

4a2from both sides gives us

(x +b

2a)2 − b2

4a2= x2 +

b

ax.



So

f(x) = a(x2 +b

ax) + c = a((x +

b

2a)2 − b2

4a2) + c

If we distribute the a into the term b2

4a2we have

f(x) = a(x +b

2a)2 − b2

4a+ c.

Now get a common denominator for the last expression:

f(x) = a(x +b

2a)2 − b2 − 4ac

4a.

This is the “standard form” for an arbitrary quadratic.(Gosh, I hope no one tries to memorize this answer! It is much easier,when given a general quadratic, to quickly complete the square to get thestandard form. If there is anything one might try to memorize, it is thegeneral solution that this result gives us – next.)



In the previous slide we completed the square on the general quadratic

f(x) = ax2 + bx + c

to turn this into “standard form”

f(x) = a(x +b

2a)2 − b2 − 4ac

4a.

Now that we have completed the square on a general quadratic, we couldask where that quadratic is zero. If we wish to solve the equation

ax2 + bx + c = 0

we might instead use the standard form and solve

a(x + b2a)2 − b2−4ac

4a = 0.

This is easy to do: Add b2−4ac4a to both sides:

a(x + b2a)2 = b2−4ac

4a ,

and divide both sides by a

(x + b2a)2 = b2−4ac

4a2,

and then take the square root of both sides.... (next slide)Smith (SHSU) Elementary Functions 2013 26 / 35


Review: ax2 + bx + c = 0 implies

a(x + b2a)2 − b2−4ac

4a = 0.

a(x + b2a)2 = b2−4ac

4a ,

(x + b2a)2 = b2−4ac

4a2,

and then take the square root of both sides keeping in mind that we wantboth the positive and negative square roots:

x + b2a = ±

√b2−4ac4a2

.

The denominator on right side can be simplified

x + b2a = ±

√b2−4ac2a .

Finally solve for x by subtracting b2a from both sides

x = − b2a ±

√b2−4ac2a

and combine the right hand side using the common denominator:

x = −b±√b2−4ac2a .

This is the quadratic formula and this expression is worth memorizing!



The general solution to the quadratic equation

ax2 + bx + c = 0

is

x =−b±

√b2 − 4ac

2a. (2)

Definition. The expression b2 − 4ac under the radical sign in equation 2 iscalled the discriminant of the quadratic equation and is sometimesabbreviated by a Greek letter capitalized delta: ∆.

With this notation, the quadratic formula says that the general solution tothe quadratic equation ax2 + bx + c = 0 is

x =−b±

√∆

2a.


In the next presentation, we examine equations of circles.

(END)



Lecture 2.1d, Equations of circles

Dr. Ken W. Smith


2013


The equation of a circle

We digress for a moment from our study of polynomials to consideranother important quadratic equation which appears often in calculus.This is an equation in which both x and y are squared, the equation for acircle.

Suppose we have a circle centered at the point (a, b) with radius r.

Let (x, y) be a point on the circle.We can draw a right triangle with short sides of lengths (x− a) and(y − b) and hypotenuse of length r.



By the Pythagorean Theorem,

(x− a)2 + (y − b)2 = r2. (3)

This is the general equation for a circle.



If we are given a quadratic equation in which x2 and y2 both occurtogether with coefficient 1 then we can recover the equation for a circle bycompleting the square. For example, suppose we are given the equation

x2 + 4x + y2 + 2y = 6.

We can complete the square on x2 + 4x, rewriting that as (x + 2)2 − 4and complete the square on y2 + 2y writing (y + 1)2 − 1. So our equationbecomes

(x + 2)2 − 4 + (y + 1)2 − 1 = 6 or

(x + 2)2 + (y + 1)2 = 11.

This is an equation for a circle with center (−2,−1) and radius√

11.Smith (SHSU) Elementary Functions 2013 33 / 35

One more example

Let’s do one more example.Find the center and radius of the circle with equation given by

x2 + 6x + y2 − 8y = 0

We begin by completing the square on x2 + 6x. Note thatx2 + 6x + 9 = (x + 3)2 so x2 + 6x = (x + 3)2 − 9.

We should also complete the square on y2 − 8y. Sincey2 − 8y + 16 = (y − 4)2 then y2 − 8y = (y − 4)2 − 16.Substituting, our equation for the circle becomes

(x + 3)2 − 9 + (y − 4)2 − 16 = 0

So, moving constants to the right side,

(x + 3)2 + (y − 4)2 = 9 + 16

or

(x + 3)2 + (y − 4)2 = 25

This is the equation of a circle with center (−3, 4) and radius r = 5.Smith (SHSU) Elementary Functions 2013 34 / 35

Summary

We have used the concept of “completing the square” to

1 find the standard form for a quadratic polynomial,

2 identify the transformations that turn the general parabola y = x2

into any other parabola,

3 create the quadratic formula,

4 and briefly analyze the equation of a circle.

In the next presentation, we explore polynomials in general.

(END)


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