2.1 Polynomials Quadratic Functions

Quadratic functions

Elementary Functions In the last lecture we studied polynomials of simple form f(x) = mx + b. Part 2, Polynomials Now we move on to a more interesting case, polynomials of degree 2, the Lecture 2.1a, Quadratic Functions quadratic polynomials.

Quadratic functions have form f(x) = ax2 + bx + c where a, b, c are real Dr. Ken W. Smith numbers and we will assume a 6= 0.

Sam Houston State University The graph of a quadratic polynomial is a parabola.

2013

Smith (SHSU) Elementary Functions 2013 1 / 35 Smith (SHSU) Elementary Functions 2013 2 / 35 Quadratic functions Quadratic functions

The graph of a quadratic polynomial is a parabola. Here is the standard parabola y = x2. 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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The standard form for a quadratic polynomial is The turning point (0, 0) on the parabola y = x2 is called the vertex of the f(x) = a(x − h)2 + k (1) parabola; it is moved by these transformations to the new vertex (h, k). From our understanding of transformation, beginning with the simple y = x2, If we are given the equation f(x) = ax2 + bx + c for a quadratic function, if we shift that parabola to the right h units, we can change the function into the above form by ﬁrst factoring out the (replacing x by x − h) term a so that

2 2 b c stretch it vertically by a factor of a f(x) = ax + bx + c = a(x + a x + a ). (multiplying on the outside of the parentheses by a)

2 b c and then shift the parabola up k units, We then complete the square on the expression x + a x + a (adding k on the outside of the parentheses) Let me explain completing the square.... we will have the graph for y = a(x − h)2 + k.

Smith (SHSU) Elementary Functions 2013 5 / 35 Smith (SHSU) Elementary Functions 2013 6 / 35 Completing the square Completing the square – the main idea

A major tool for solving quadratic equations is to turn a quadratic into an Here is a test of our understanding of the simple equation 2 2 2 expression involving a sum of a square and a constant term. (x + A) = x + 2Ax + A . Fill in the blanks: 1 (x + 2)2 = x2 +4 x + 4 This technique is called “completing the square.” 2 (x + 5)2 = x2 + 10x + 25 Recall that if we square the linear term x + A we get 2 2 (x + A)2 = x2 +2 Ax + A2. 3 (x + 1) = x +2 x + 1

4 (x + − 1)2 = x2−2x + 1 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 5 (x + − 3)2 = x2−6x + 9 answer. 7 49 6 (x + )2 = x2 +7 x + In the expansion of (x + A)(x + A) we summed Ax twice. 2 4

Smith (SHSU) Elementary Functions 2013 7 / 35 Smith (SHSU) Elementary Functions 2013 8 / 35 Completing the square – the main idea Completing the square – the main idea

Applying this idea. If one understands this simple idea, that we can predict the square by Since x2 − 6x is the beginning of looking at the coeﬃcient of x, then we can rewrite any quadratic polynomial into an expression involving a perfect square. x2 − 6x + 9 = (x − 3)2

For example, since then x2 + 4x + 4 = (x + 2)2 x2 − 6x = (x − 3)2 − 9 then a polynomial that begins x2 + 4x must involve (x + 2)2. and so x2 − 6x + 2 = (x − 3)2 − 7, For example: x2 + 4x = (x + 2)2 − 4 x2 − 6x + 10 = (x − 3)2 + 1,

x2 + 4x + 7 = (x + 2)2 + 3 x2 − 6x − 3 = (x − 3)2 − 12,

x2 + 4x − 5 = (x + 2)2 − 9

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This is useful if we are trying to put an equation y = f(x) of a quadratic Since x2 − 3x is the beginning of into the standard form y = a(x − h)2 + k. 9 3 x2 − 3x + = (x − )2 In the next presentation, we apply the concept of “completing the square” 4 2 to put a quadratic into standard form and then we will use that form to then develop the quadratic formula. 3 9 x2 − 3x = (x − )2 − . 2 4 (END)

Smith (SHSU) Elementary Functions 2013 11 / 35 Smith (SHSU) Elementary Functions 2013 12 / 35 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 Elementary Functions Part 2, Polynomials x2 + 4x − 5 = (x + 2)2 − 9 Lecture 2.1b, Applications of completing the square

x2 − 6x + 9 = (x − 3)2 Dr. Ken W. Smith

2 2 Sam Houston State University x − 6x = (x − 3) − 9.

2013 x2 − 6x + 2 = (x − 3)2 − 7,

x2 − 6x + 10 = (x − 3)2 + 1,

2 2 Smith (SHSU) Elementary Functions 2013 13 / 35 Smith (SHSU) x − 6x Elementary− 3 = ( Functionsx − 3) − 12. 2013 14 / 35 Completing the square – the main idea Completing the square – the main idea

For example, if y = x2 + 4x + 7

Since x2 − 3x is the beginning of is the equation of a parabola then we can complete the square, writing x2 + 4x = (x + 2)2 − 4 and so 9 3 2 2 2 x2 − 3x + = (x − )2 x + 4x + 7 =( x + 2) − 4+7 = (x + 2) + 3. Our equation is now 4 2 y = (x + 2)2 + 3. then 3 9 x2 − 3x = (x − )2 − . This is the standard form for the quadratic. We see that the vertex of the 2 4 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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An extra step occurs if the coeﬃcient of x2 is not 1.

How do we complete the square on something like 2x2 + 8x + 15? What is the vertex of the parabola with equation y = 2x2 + 8x + 15? Once again we focus on the ﬁrst two terms, 2x2 + 8x. Factor out the coeﬃcient of x2 so that Solution. Complete the square to write

2 2 2x2 + 8x = 2(x2 + 4x). y = 2x + 8x + 15 = 2(x + 2) + 7.

2 Now complete the square inside the parenthesis so that The parabola with equation y = 2(x + 2) + 7 has vertex (−2, 7). x2 + 4x = (x2 + 4x + 4) − 4 = (x + 2)2 − 4. Therefore To transform the parabola y = x2 to the graph of y = 2(x + 2)2 + 7 2 2 2 2 2x + 8x = 2(x + 4x) = 2[(x + 2) − 4] = 2(x + 2) − 8. 1 ﬁrst shift y = x2 left by 2 units 2 then stretch the graph vertically by a factor of 2 If y = 2x2 + 8x + 15 then completing the square gives 3 and then slide the graph up 7 units. y = 2x2 + 8x + 15 = 2(x + 2)2 − 8 + 15 = 2(x + 2)2 + 7 .

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Find the standard form for a parabola with vertex (2, 1) passing through (4, 5). Find the vertex of the parabola with graph given by the equation

Solution. A parabola with vertex (2, 1) passing through (4, 5) has y = 2x2 − 12x + 4. 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. Solution. Completing the square, we see that

2 2 2 By plugging in the point (4, 5) we see that 2x − 12x + 4 = 2(x − 6x) + 4 = 2((x − 3) − 9) + 4

5 = a(4 − 2)2 + 1 = 2(x − 3)2 − 18 + 4 = 2(x − 3)2 − 14. and so a = 1.

So our answer is So the vertex is (3, −14). y =1( x − 2) 2 +1

Smith (SHSU) Elementary Functions 2013 19 / 35 Smith (SHSU) Elementary Functions 2013 20 / 35 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. Elementary Functions Solution. Since the standard form for the parabola is y = 2(x − 3)2 − 14 Part 2, Polynomials (from the previous slide) then we must do the following transformations, in Lecture 2.1c, The Quadratic Formula this order: 1 Shift y = x2 right by 3. 2 Stretch the graph by a factor of 2 in the vertical direction. Dr. Ken W. Smith

3 Shift the graph down by 14. Sam Houston State University

In the next presentation we discuss the quadratic formula 2013 (END)

Smith (SHSU) Elementary Functions 2013 21 / 35 Smith (SHSU) Elementary Functions 2013 22 / 35 The quadratic formula The quadratic formula, continued

We complete the square on We obtain a nice general formula for solutions to a quadratic equation if 2 we complete the square on a general, arbitrary quadratic function. Let’s f(x) = ax + bx + c see how this works. by factoring a from the ﬁrst two terms, b Consider the general quadratic function f(x) = a(x2 + x) + c. a 2 f(x) = ax + bx + c 2 b 2 b Then we complete the square on x + a x. Since x + a x is the beginning where a, b and c are unknown constants. b 2 of the expression for (x + 2a ) we compute b b b2 We can put this function into standard form by completing the square. (x + )2 = x2 + x + . 2a a 4a2 First we factor out a from the ﬁrst two terms,

2 b Subtracting b from both sides gives us f(x) = a(x2 + x) + c. 4a2 a b b2 b (x + )2 − = x2 + x. 2a 4a2 a

Smith (SHSU) Elementary Functions 2013 23 / 35 Smith (SHSU) Elementary Functions 2013 24 / 35 The quadratic formula The quadratic formula

So In the previous slide we completed the square on the general quadratic b b b2 f(x) = a(x2 + x) + c = a((x + )2 − ) + c f(x) = ax2 + bx + c a 2a 4a2 to turn this into “standard form” 2 2 b b 2 b − 4ac If we distribute the a into the term 2 we have f(x) = a(x + ) − . 4a 2a 4a b b2 f(x) = a(x + )2 − + c. Now that we have completed the square on a general quadratic, we could 2a 4a ask where that quadratic is zero. If we wish to solve the equation 2 Now get a common denominator for the last expression: ax + bx + c = 0 2 we might instead use the standard form and solve b 2 b − 4ac f(x) = a(x + ) − . b 2 b2−4ac 2a 4a a(x + 2a ) − 4a = 0. 2 This is easy to do: Add b −4ac to both sides: This is the “standard form” for an arbitrary quadratic. 4a b 2 b2−4ac (Gosh, I hope no one tries to memorize this answer! It is much easier, a(x + 2a ) = 4a , when given a general quadratic, to quickly complete the square to get the and divide both sides by a standard form. If there is anything one might try to memorize, it is the 2 (x + b )2 = b −4ac , general solution that this result gives us – next.) 2a 4a2 Smith (SHSU) Elementary Functions 2013 25 / 35 and thenSmith take (SHSU) the square root ofElementary both Functions sides.... (next slide) 2013 26 / 35 The quadratic formula The quadratic formula

Review: ax2 + bx + c = 0 implies The general solution to the quadratic equation b 2 b2−4ac a(x + 2a ) − 4a = 0. ax2 + bx + c = 0 b 2 b2−4ac a(x + 2a ) = 4a , is √ 2 (x + b )2 = b −4ac , −b ± b2 − 4ac 2a 4a2 x = . (2) 2a and then take the square root of both sides keeping in mind that we want both the positive and negative square roots: Deﬁnition. The expression b2 − 4ac under the radical sign in equation 2 is b q b2−4ac x + 2a = ± 4a2 . called the discriminant of the quadratic equation and is sometimes The denominator on right side can be simpliﬁed abbreviated by a Greek letter capitalized delta: ∆. √ b b2−4ac x + 2a = ± 2a . With this notation, the quadratic formula says that the general solution to b the quadratic equation ax2 + bx + c = 0 is Finally solve for x by subtracting 2a from both sides √ 2 √ x = − b ± b −4ac −b ± ∆ 2a 2a x = . and combine the right hand side using the common denominator: 2a √ −b± b2−4ac Smith (SHSU) x Elementary= Functions2a . 2013 27 / 35 Smith (SHSU) Elementary Functions 2013 28 / 35 This is the quadratic formula and this expression is worth memorizing! Elementary Functions In the next presentation, we examine equations of circles. Part 2, Polynomials Lecture 2.1d, Equations of circles (END) Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 29 / 35 Smith (SHSU) Elementary Functions 2013 30 / 35 The equation of a circle 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.

By the Pythagorean Theorem,

(x − a)2 + (y − b)2 = r2. (3)

This is the general equation for a circle.

Smith (SHSU) Elementary Functions 2013 31 / 35 Smith (SHSU) Elementary Functions 2013 32 / 35 The equation of a circle 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.

If we are given a quadratic equation in which x2 and y2 both occur We should also complete the square on y2 − 8y. Since together with coeﬃcient 1 then we can recover the equation for a circle by y2 − 8y + 16 = (y − 4)2 then y2 − 8y = (y − 4)2 − 16. completing the square. For example, suppose we are given the equation Substituting, our equation for the circle becomes x2 + 4x + y2 + 2y = 6. (x + 3)2 − 9+( y − 4)2 − 16=0 We can complete the square on x2 + 4x, rewriting that as (x + 2)2 − 4 So, moving constants to the right side, 2 2 and complete the square on y + 2y writing (y + 1) − 1. So our equation (x + 3)2 + (y − 4)2 = 9 + 16 becomes or (x + 2)2 − 4 + (y + 1)2 − 1 = 6 or (x + 3)2 + (y − 4)2 = 25 (x + 2)2 + (y + 1)2 = 11. Smith (SHSU) Elementary Functions √2013 33 / 35 This isSmith the (SHSU) equation of a circleElementary with center Functions (−3, 4) and radius r2013= 5. 34 / 35 This is an equation for a circle with center (−2, −1) and radius 11. 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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