Quadratic functions Elementary Functions In the last lecture we - - PowerPoint PPT Presentation

Elementary Functions

Part 2, Polynomials Lecture 2.1a, Quadratic Functions

• Dr. Ken W. Smith

Sam Houston State University

2013

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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, the quadratic polynomials. Quadratic functions have form f(x) = ax2 + bx + c where a, b, c are real numbers and we will assume a = 0. The graph of a quadratic polynomial is a parabola.

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Quadratic functions

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

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Quadratic functions

All of the graphs of quadratic functions can be created by transforming the parabola 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 quadratic function. 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.)

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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 simple y = 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.

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Quadratic functions

The turning point (0, 0) on the parabola y = x2 is called the vertex of the parabola; 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 the term a so that f(x) = ax2 + bx + c = a(x2 + b

ax + c a).

We then complete the square on the expression x2 + b

ax + c a

Let me explain completing the square....

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Completing the square

A major tool for solving quadratic equations is to turn a quadratic into an expression 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 expression but the expression 2Ax, the “cross-term” is also a critical part of our answer. In the expansion of (x + A)(x + A) we summed Ax twice.

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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

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Completing the square – the main idea

Applying this idea. If one understands this simple idea, that we can predict the square by looking at the coefficient of x, then we can rewrite any quadratic polynomial into an expression involving a perfect square. For example, since x2 + 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

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Completing the square – the main idea

Since x2 − 6x is the beginning of x2 − 6x + 9 = (x − 3)2 then x2 − 6x = (x − 3)2 − 9 and so x2 − 6x + 2 = (x − 3)2 − 7, x2 − 6x + 10 = (x − 3)2 + 1, x2 − 6x − 3 = (x − 3)2 − 12,

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Completing the square – the main idea

Since x2 − 3x is the beginning of x2 − 3x + 9 4 = (x − 3 2)2 then x2 − 3x = (x − 3 2)2 − 9 4.

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Completing the square – the main idea

This is useful if we are trying to put an equation y = f(x) of a quadratic into 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 to develop the quadratic formula. (END)

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Elementary Functions

Part 2, Polynomials Lecture 2.1b, Applications of completing the square

• Dr. Ken W. Smith

Sam Houston State University

2013

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Completing the square – the main idea

In the last lecture we discussed quadratic functions and introduced the concept 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.

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Completing the square – the main idea

Since x2 − 3x is the beginning of x2 − 3x + 9 4 = (x − 3 2)2 then x2 − 3x = (x − 3 2)2 − 9 4.

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Completing the square – the main idea

For example, if y = x2 + 4x + 7 is the equation of a parabola then we can complete the square, writing x2 + 4x = (x + 2)2 − 4 and so x2 + 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 the parabola is (−2, 3). We can transform the graph of y = x2 into the graph of y = (x + 2)2 + 3 by shifting the graph of y = x2 two units to the left and then shifting the graph up 3 units.

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Completing the square – the main idea

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 that x2 + 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 .

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Completing the square – the main idea

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.

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Completing the square – the main idea

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) has

standard form y = a(x − h)2 + k where (h, k) is the coordinate of the vertex. 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 is y = 1 (x − 2 )2 + 1

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Completing the square – the main idea

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).

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Completing the square – the main idea

Describe, precisely, the transformations necessary to move the graph of y = 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, in this 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)

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Elementary Functions

Part 2, Polynomials Lecture 2.1c, The Quadratic Formula

• Dr. Ken W. Smith

Sam Houston State University

2013

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The quadratic formula

We obtain a nice general formula for solutions to a quadratic equation if we complete the square on a general, arbitrary quadratic function. Let’s see 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.

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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 + b

• ax. Since x2 + b

ax is the beginning

• f the expression for (x + b

2a)2 we compute

(x + b 2a)2 = x2 + b ax + b2 4a2 . Subtracting

b2 4a2 from both sides gives us

(x + b 2a)2 − b2 4a2 = x2 + b ax.

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The quadratic formula

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 4a2 we 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 the standard form. If there is anything one might try to memorize, it is the general solution that this result gives us – next.)

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The quadratic formula

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 could ask 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 + b

2a)2 − b2−4ac 4a

= 0. This is easy to do: Add b2−4ac

4a

to both sides: a(x + b

2a)2 = b2−4ac 4a

, and divide both sides by a (x + b

2a)2 = b2−4ac 4a2 ,

and then take the square root of both sides.... (next slide)

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The quadratic formula

Review: ax2 + bx + c = 0 implies a(x + b

2a)2 − b2−4ac 4a

= 0. a(x + b

2a)2 = b2−4ac 4a

, (x + b

2a)2 = b2−4ac 4a2 ,

and then take the square root of both sides keeping in mind that we want both the positive and negative square roots: x + b

2a = ±

• b2−4ac

4a2 .

The denominator on right side can be simplified x + b

2a = ± √ b2−4ac 2a

. Finally solve for x by subtracting

b 2a from both sides

x = − b

2a ± √ b2−4ac 2a

and combine the right hand side using the common denominator: x = −b±

√ b2−4ac 2a

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

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The quadratic formula

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 is

called the discriminant of the quadratic equation and is sometimes abbreviated by a Greek letter capitalized delta: ∆. With this notation, the quadratic formula says that the general solution to the quadratic equation ax2 + bx + c = 0 is x = −b ± √ ∆ 2a .

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In the next presentation, we examine equations of circles. (END)

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Elementary Functions

Part 2, Polynomials Lecture 2.1d, Equations of circles

• Dr. Ken W. Smith

Sam Houston State University

2013

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The equation of a circle

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

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The equation of a circle

By the Pythagorean Theorem, (x − a)2 + (y − b)2 = r2. (3) This is the general equation for a circle.

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The equation of a circle

If we are given a quadratic equation in which x2 and y2 both occur together with coefficient 1 then we can recover the equation for a circle by completing 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 − 4 and complete the square on y2 + 2y writing (y + 1)2 − 1. So our equation becomes (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.

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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 that x2 + 6x + 9 = (x + 3)2 so x2 + 6x = (x + 3)2 − 9. We should also complete the square on y2 − 8y. Since y2 − 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

• r

(x + 3)2 + (y − 4)2 = 25 This is the equation of a circle with center (−3, 4) and radius r = 5.

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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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