6 Linear and Quadratic Functions

Transcription

1 6 Linear and Quadratic Functions 6. Prologue Definition 4 in Book I of Euclid s Elements (3rd centur BC) reads: A straight line is a line that lies evenl with the points on itself. No one knows for sure what eactl he meant b this mabe the meaning got lost in translation. Perhaps he meant that if ou aim from one point to another, all the points in between fall on that line and no points stick out. In an case, we know what a straight line is. Or do we? A stretched piece of string? A ra of light? This is probabl the most deceptive intuitive notion. Einstein s theor of relativit states that ras of light are actuall bent b gravit. In general, our faith in Euclidean geometr has been shaken b more recent (8th centur) models of non-euclidean geometries, in which there eists more than one line through a given point that is parallel to a given line, or in which no parallel lines eist at all. What if our space is actuall curved? Here we ll avoid these mind-bending considerations and stick with the Euclidean point of view. Copright 00 b Sklight Publishing We define a straight line on the Cartesian plane as a graph of a linear relation p + q = C. If q 0, this relation is a function of the form = m + b, where p C m = and b =. A function = m + b is called a linear function. In q q Sections <...>-<...>, we will discuss the properties of linear relations and functions. A function given b the formula f ( ) = a + b+ c, where a, b, and c are constants and a 0, is called a quadratic function. Its graph is a parabola, a curve with man wonderful properties. We will discuss the properties of quadratic functions and their graphs in Sections <...>-<...>. 6. Linear Relations A relation from to of the form p + q = C, where p, q, and C are constants (and at least one of p and q is not equal to 0) is called a linear relation. The graph of a linear relation in the Cartesian plane is a straight line.

2 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS It is not ver hard to prove that this definition of a straight line is consistent with the postulates of Euclidean geometr. For eample, Postulate states that we can draw a straight line through an two points. Indeed, the points (, ) and (, ) lie on the graph of p + q = C, where p = ( ), q= ( ), and C = (see Question <...> in the eercises). It is also not ver hard to prove that there is onl one line that can be drawn through an two given points. A straight line is an abstraction: our intuition (or our education) tells us that we can align an imaginar ruler, called a straightedge, with an two points, and draw a straight line through them with an infinitel thin pencil. Eample Find a linear relation that defines a straight line through the points (, 5) and (, 4). Solution Copright 00 b Sklight Publishing Using the above formulas, p = ( ), q= ( ), and C =, we get: p= = 4 5=, q= ( ) = ( ) =, C = = 4 5= 3. Thus, the relation + = 3 or = 3 defines the line through (, 5) and (, 4). B our definition, the -ais is a straight line, because it is the graph of the relation 0 + = 0 = 0. In general, if p = 0, the graph is the horizontal line C =. Similarl, the -ais is a straight line, because it is the graph of q C + 0 = 0 = 0. In general, if q = 0, the graph is the vertical line =. p Two straight lines that do not intersect (have no common points) are called parallel. Clearl the graphs of p + q = C and p + q = C cannot share a point when C and C are different, so these lines are parallel. If p and q are fied, and we var C, we get a famil of parallel lines (Figure 6-). One of them, the graph of p + q = 0, passes through the origin.

3 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 3 O p+ q = C 3 p+ q = C p+ q = C p + q = 0 Figure 6-. The graphs of p + q = C for different values of C produce a famil of parallel lines Copright 00 b Sklight Publishing This model of Euclidean geometr is also consistent with Euclid s fifth postulate, which is equivalent to the statement that given a straight line l and a point P outside it, we can draw precisel one line through P parallel to l: l P If a straight line is neither vertical nor horizontal, that is, p 0 and q 0, and it doesn t pass through the origin, that is C 0, then it has one -intercept, a 0 and one -intercept, b 0 : b O a

4 4 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS An equation of such a line can be written as + =, where a b C = ). p =, a q =, and b This form of the equation of the line is called the intercept form. Eample Sketch a straight line through (, 0) and (0, ) and determine its equation. Solution - - O - Copright 00 b Sklight Publishing The -intercept is and the -intercept is. Using the intercept form, we immediatel find the equation: + = or + =. Eample 3 Sketch the graph of + = 3. Solution We know from Eample what the graph of + = looks like. The graph + = 3 is a line parallel to the line + =, but its -intercept is 3, not :

5 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 5 - O 3 - To solve this problem from scratch, set to 0 to get the -intercept, 3; then set to 0 to get the -intercept, 3, then connect the two intercept points with the straight line. An line p + q = C is perpendicular to an line ( q ) + p = C. Recall from Chapter <...> that two vectors are perpendicular to each other if and onl if their dot product is equal to 0. Also recall that the vector v, which represents the difference of the vectors (, ) and (, ), v = (, ), is parallel to the line that passes through the points (, ) and (, ) : Copright 00 b Sklight Publishing O (, ) v (, ) Suppose an equation of this line is p + q = C. If u = ( p, q), then the dot product u v = p( ) + q( ) = ( p + q) ( p + q) = C C = 0. This means u is perpendicular to v. Thus a line p + q = C is perpendicular to the vector u = ( p, q). We can see it graphicall, too. For eample: (, ) + = 3 - O 3 -

6 6 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS If we rotate u b 90 degrees, we get the vector u = ( q, p) : it has the same length as u and is perpendicular to it. Indeed, their dot product, u u = p( q) + qp= 0. Eample 4 Find an equation of the line that contains the point (,3) and is perpendicular to the line =. Solution The equation has the form ( 5) + 3 = C. For the point (,3) to be on that line, we must have ( 5)( ) + 3 3= C C = 4. The answer is 5+ 3 = 4. Eercises. When and where did Euclid live? How man books make up Euclid s Elements? Copright 00 b Sklight Publishing In Questions -5, write an equation p + q = C for the straight line that passes through the two given points.. (-, -) and (3, 5) 3. (, 4) and (-, -) 4. (, ) and (, 5) 5. (, -) and (-3, -) In Questions 6-9, write an equation for the straight line that passes through the given point and is parallel to the given line. 6. Point (-3, ) and line + = Point (, 5) and line = Point (4, -) and line + 3 = 7.

7 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 7 9. Point (-, ) and line =. In Questions 0-3, write an equation for the straight line through the two given points in the intercept form, + =. a b 0. (0, 3) and (4, 0).. (0, -) and (3, 0).. (4, 0) and (0, -). 3. (-3, ) and (, -3). Hint: First write p + q = C, then find the intercepts. In Questions 4-7, write an equation for the straight line that passes through the given point and is perpendicular to the given line. 4. Point (0, 0) and line + =. Copright 00 b Sklight Publishing 5. Point (6, -3) and line =. 6. Point (3, -3) and line = Point (-, ) and line =. 8. Prove that the line p + q = C, where p = ( ), q= ( ), and C = contains the points (, ) and (, ). 9. Prove that onl one straight line passes through the points (, ) and (, ). Hint: First prove for the case where one of the points is the origin; then, reduce the general case to this special case. 0. The lines 4+ 3 = 5 and 4+ 3 = 75 are parallel. What is the distance between them? Hint: Draw a perpendicular line through the origin and find the distance between its intersections with the given lines.

8 8 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS. Consider the surface of a sphere. Let us define a straight line on it as a great circle (that is, a circle on the surface of the sphere centered at the center of the sphere). Is Euclid s First Postulate (that there is a unique straight line through an two points) true in this geometr? What about the Fifth Postulate (equivalent to the statement that for an line l and a point P outside of it there eists precisel one line through P parallel to l)? 6.3 Linear Functions A linear relation p + q = C in which q 0 can be rewritten as = m + b, where p C m = and b =. q q The formula f ( ) = m+ b defines a linear function. The natural domain of this function is all real numbers. m is called the slope of the line and b is its -intercept (Figure 6-). Copright 00 b Sklight Publishing b =m+b Δ θ Δ ( delta- ) ( delta- ) Δ m= ( m= tanθ ) O Δ Figure 6-. The slope m of a line = m + b is equal to Δ ( delta ) over Δ ( delta ) (equal to the tangent of the angle the line makes with the -ais) The = m + b form of an equation for a straight line is called the slope-intercept form, where m is the slope and b is the -intercept.

9 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 9 Some teachers use the phrase rise over run to help their students understand how the slope is measured. But the rise Δ ( delta ) can actuall be a fall when the slope is negative ( Δ < 0 ): =m+b Δ ( delta- ) b O θ Δ ( delta- ) π In an case, the slope is alwas equal to tanθ ; when 90 < θ < 80 ( < θ < π), tanθ is negative. A constant function f ( ) = b is a special case of a linear function; its graph is a horizontal line with the slope m = 0. Copright 00 b Sklight Publishing The slope of a vertical line is undefined (infinit). The line = m passes through the origin. If the slope m is positive, the line is located in the first and third quadrants: = m m > 0 If the slope is negative, the line is located in the second and fourth quadrants: =m m < 0

10 0 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS Eample Rewrite the equation 3 = 7 in slope-intercept form. Solution 3 = 7 3 = = = Eample What is the slope of the line + =? Solution + = = +, so the slope is. Copright 00 b Sklight Publishing Eample 3 Sketch the graph of =. Solution The -intercept is and the slope is :

11 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS If a line passes through the points (, ) and (, ) and it is not vertical (that is, ), then its slope m = (Figure 6-3). O (, ) (, ) θ m = Figure 6-3. The slope m of a line through the points (, ) and (, ) is m = Copright 00 b Sklight Publishing So, if (, ) is a point on a non-vertical line, then for an other point (, ) on that line = m m = ( ) or m( ) = +. This form of an equation for a line is called the point-slope form. Eample Write an equation of a line shown on the graph above.

12 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS Solution The line passes through the point (, ) and has a slope of. Its equation can be written as = ( ) = + = Eample 5 Write an equation of the line that passes through the points (,3) and (, 5). Solution ( 5) m = = = =. Using the point-slope form, ( ) = ( ( ) ) + 3 = ( + ) + 3= + 3 = Copright 00 b Sklight Publishing Different forms of an equation for a straight line are convenient in different situations. To summarize: Standard form: p + q = C Slope-intercept form: = m + b Point-slope form: = m ( ) or = m ( ) + Two-point form, : = m ( ) +, where m = Horizontal line: = b Vertical line: = a

13 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 3 Recall from Section <...> that the lines p + q = C and q + p = C are p perpendicular to each other. The slope of the first line is = m ; the slope of the q second line is ( q) q = =. Therefore, p p m an two lines = m+ b and = + b are perpendicular to each m other. In other words, if two lines with slopes m and m are perpendicular to each other, then m =. m Eample 6 Copright 00 b Sklight Publishing Write an equation of the line that passes through the origin and is perpendicular to the line from Eample 5. Solution In Eample 5 we found that the slope of the line that passes through the points 8 (,3) and (, 5) is m =. The slope of the perpendicular line is = =. The answer is =. m 8 8 8

14 4 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS Eercises In Questions -4 write the slope-intercept equation of the line shown in the graph Copright 00 b Sklight Publishing In Questions 5-8 sketch b hand the graph of the given linear function and label the -intercept. 5. = 6. = 3 7. = = +

15 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 5 In Questions 9- write an equation of the line that passes through the given point and is parallel to the given line. 9. Point (0, 0), line = Point (, ), line = 3.. Point (, ), line = Point (0, 3), line =. In Questions 3-6 find the slope of the line that passes through the two given points, write its equation, sketch its graph, and label its -intercept. 3. ( 3, ) and (3,) 4. (, 0) and (5, 3) 5. (, ) and ( 4,) Copright 00 b Sklight Publishing 6. ( 3,3) and (, 3) In Questions 7-0 write an equation of the line that passes through the given point and is perpendicular to the given line. 7. Point (0, 0), line = Point (,), line = Point ( 4, ), line = Point (, ), line = What is the distance from the origin to the line 3 5 = +?. Given a triangle ABC with the vertices A (, 3), B(, 3), and C( 4, ), write an equation of the line that contains the altitude from A to BC.

16 6 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 3. The point P (3, 4) lies on a circle centered at the origin. What is the -intercept of the line tangent to the circle at P? Hint: A tangent line to a circle is perpendicular to the radius at the point of tangenc. 4. In Question <...> on page <...> we introduced the golden rectangle: it is a rectangle of such a shape that if ou cut off a square from it, the remaining rectangle has the same shape (aspect ratio) as the original one: A φ B ϕ = ϕ O M φ- C Copright 00 b Sklight Publishing The Greek letter ϕ (phi, pronounced as fie ) is commonl used to denote the golden ratio. Suppose a golden rectangle is placed in the Cartesian plane with O at the origin, A at (0, ), and C at (φ, 0). Write the equations of the lines AC and BM and prove that the are perpendicular to each other: A O 5. In linear algebra, the branch of mathematics that studies vector spaces and linear transformations on them, a function is called linear if, for an and, f ( + ) = f ( ) + f ( ) and for an real number k and for an, f ( k) = kf ( ). Are an of the linear functions f ( ) = a+ b on (which is a one-dimensional vector space) also linear in the linear algebra sense? Which ones? M B C

17 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS Parabolas A function from to given b a formula a 0, is called a quadratic function. f ( ) a b c = + +, where The simplest quadratic function is f ( ) =. The graph of this function is a curve called a parabola (Figure 6-4). As we will see in the net section, the graph of an quadratic function is a parabola, perhaps stretched or shrunk, shifted and/or reflected over the -ais. = 0 Copright 00 b Sklight Publishing Figure 6-4. A parabola The parabola in Figure 6-4 is smmetric with respect to the -ais. An parabola is smmetric with respect to one line, called the ais of the parabola. The point where its ais intersects the parabola is called the verte of the parabola. The verte of the parabola in Figure 6-4 is the origin. The parabola in Figure 6-4 appears to have a horizontal tangent line at the verte. How do we know that the parabola does not have a corner or even a cusp at the verte? Something like this:??

18 8 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS To show that the graph of = indeed has a tangent line at the verte, let s take a point (, ) on the parabola and draw a line through that point and the origin: (, ) A line passing through two points of a curve is called a secant line. The slope of this line is = =. If we move the point (, ) along the parabola toward the origin, becomes smaller, and the slope of the secant line gets closer and closer to 0. We can sa that the limit of the secant line, as approaches 0, is the horizontal tangent line through the origin. Copright 00 b Sklight Publishing A parabola is a beautiful shape. First, if we take a cone and cut it with a plane parallel to one side of the cone, the shape of the cross-section is a parabola (Figure 6-5). In fact, the name parabola means a conic section in ancient Greek; the name is due to the Greek geometer Apollonius (about 6 BC - about 90 BC) who studied conic sections. Figure 6-5. A cross-section of a cone with a plane parallel to its side is parabola Second, an parabola is the locus of points (the set of all points) equidistant from a certain straight line, called the directri, and a certain point, called the focus of the

19 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 9 parabola (Figure 6-6). The focus lies on the ais of the parabola, and the directri is perpendicular to the ais. F l Figure 6-6. The locus of points equidistant from a given point F and a line l is a parabola. l is called the directri of the parabola and F is called its focus. Third, if ou make a mirror in the shape of a parabola, an ra of light emitted from the focus of the parabola will be reflected in the direction parallel to the ais (Figure 6-7). No wonder parabolicall shaped reflectors are used in the headlights of cars and in spotlights. Copright 00 b Sklight Publishing F Figure 6-7. A parabola reflects all the ras from its focus F in the same direction, parallel to the ais of the parabola. Eample Find the focus and the directri for the graph of =.

20 0 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS Solution The focus of this parabola lies on the -ais, and the directri is a horizontal line. Suppose the focus is at (0, d). Then the directri must be the line = d, because the verte (0, 0) must be equidistant from the focus and the directri: = (, ) =-d 0 d d +d Copright 00 b Sklight Publishing We need to find d. Let us take an point (, ) on the parabola. It must be equidistant from the focus and the directri. The distance from the point (, ) to the directri is + d. The distance from the point (, ) to the focus (b the distance formula) is ( 0) ( d) + = + ( d). We must have: + d = + ( d) ( + d) = + ( d) (recall that ( a± b) = a ± ab+ b ) + d + d = + d + d ( and d terms cancel out) d = d 4d = Since (, ) is on the parabola, = 4d = d =. The focus is 0, 4 4 and the directri is =. 4 Eample Find an equation for the parabola that is the locus of points equidistant from the -ais and the point (0, 4).

21 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS Solution Since the focus lies on the -ais, the -ais is the ais of the parabola. The verte of the parabola lies halfwa between the -ais and the directri, at (0, ). This means our parabola is the standard parabola =, stretched or shrunk b some factor and shifted up b units: = a +. When =, = a +, so the point (, a + ) lies on the parabola. The distance from that point to the directri is a +, and the distance from the focus is: ( 0 ) (( a ) 4) + + = ( ) + a = a 4a = a 4a+ 5 We must have: ( a+ ) = a 4a+ 5 a a a a = a= a=. 8 An equation for this parabola is = +. 8 Copright 00 b Sklight Publishing Eercises. Look up the Greek geometer Apollonius on the Internet. When and where did he live? What is his major work? In Questions -5, sketch the graph of the given quadratic function and label three points on the graph with the eact coordinates.. 3. = = = + 3 = 4 6. Show that the graph of = a is smmetrical with respect to the -ais.

22 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 7. The points,3 and (, 3) lie on the graph of = 3+ a parabola. Find the ais of this parabola, then its verte, then sketch the graph of = 3+. Hint: Since the given points are at the same horizontal level, the ais must be halfwa between them. 8. Find the -intercepts of the parabola verte, then sketch its graph. Hint: = 5 + 6, then find its ais and its = ( )( 3). 9. Find the focus and the directri of the parabola 0. Find the focus and the directri of the parabola = 3 +. = 4.. The graph of the function = is a parabola that has the same shape as the graph of =, but its ais is the line =. Find the coordinates of its verte, then its focus and directri. Copright 00 b Sklight Publishing. Write an equation for the locus of points on the Cartesian plane that are equidistant from the point (0, 3) and the line =. 3. Write an equation for the locus of points on the Cartesian plane that are equidistant from the point (0, ) and the line = 4.

23 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 3 4. A l F M D In the picture above, F is the focus of the parabola, l is its directri, the point A is on the parabola, and AD l. A is equidistant from the focus and the directri, so AF = AD. We want to show that the ra FA is reflected b the parabola in the direction parallel to the ais (and perpendicular to the directri) of the parabola. In other words, it must be reflected along DA. If AM is tangent to the parabola at A, then FAM must be equal to the angle of reflection, which is the same as DAM. Show that this is true. Hint: Show that the perpendicular bisector of FD shares onl one point with the parabola (point A) so it is tangent to the parabola. Copright 00 b Sklight Publishing 5. As we saw in Eample, the focus of the parabola = is 0, and the 4 directri is =. Find, in terms of, the slope of the line tangent to the 4 parabola at the point A(, ). Hint: see the diagram in Question 4: AM is the tangent line; AM FD; find the slope of FD, then the slope of AM.

24 4 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 6. Suppose that the points A, B, and C lie on the graph of = and have -coordinates of d,, and + d, respectivel. Show that the area of ABC does not depend on, onl on d (more precisel, the area of 3 ABC = d ). Hint: = A B C P Q R The area of ABC is equal to the area of the trapezoid PACR minus the areas of the trapezoids PABQ and QBCR. Copright 00 b Sklight Publishing 7. Archimedes (3rd centur BC) was the first to calculate the area of a segment of a parabola below a chord. He represented the area as an infinite sum of triangles inscribed in the parabola. For the first triangle, ABC, he chose the point B directl under the midpoint of AC : A C A He then similarl inscribed two triangles, one under AB, another under BC. Then four triangles under each of the four chords, and so on. Show that the area of the segment of the parabola below AC is equal to 4 of the area of 3 ABC. Hint: See Question 6. What is the area of the parabola = under the horizontal line =? B C 8. R is the region in the Cartesian plane where ; R is the region where. What is the area of R R? Hint: See Question 7.

25 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS Quadratic Functions and the Quadratic Formula In the previous section we defined a quadratic function as a function of the form f ( ) = a + b+ c, where a 0. The graph of an quadratic function is a parabola. Quadratic functions come up frequentl in science, applied mathematics, engineering, and even everda life. For eample, the trajector of a fl ball is a parabola (if we don t take into account the curvature of the earth s surface and air resistance). In man applications we are interested in the zeros (roots) of a function (When will the ball hit the ground?) or in its minimum or maimum value (When or where will the ball reach its highest point above the ground and at what altitude?). In other words, we might need to solve a quadratic equation a + b + c = 0, and/or to find the coordinates of the verte of a parabola = a + b + c. (We will sometimes sa informall the parabola = a + b+ c, meaning the graph of = a + b + c.) In this section we will derive formulas for the -coordinates of the verte of a parabola and the formula for solving a quadratic equation. Copright 00 b Sklight Publishing Let us first find the -coordinate of the ais of a parabola, which is the same as the -coordinate of its verte. Recall that a parabola is smmetric with respect to its ais. So if f ( ) = a + b+ c, f( ) = 0 and f( ) = 0, the ais of the parabola will be halfwa between and : ( v, v ) = a + b+ c = v + The graph of g( ) a b = + is the same parabola as the graph of f ( ) = a + b+ c, onl shifted verticall b c units (up or down). Therefore, these two parabolas have the same ais, and the -coordinates of their vertices are the b same. g ( ) = a +, so (0) 0 a g = and b g = a 0. The ais of the graph of g is

26 6 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS b 0 + a b the line = =. The -coordinate of the verte of this parabola, as a b well as the parabola = a + b + c, is v =. To find the -coordinate, we a simpl plug v into the function: b b v = av + bv + c = a + b + c= a a b b b + c = + c 4a a 4a a b 4a b + c= a The verte of the parabola = a + b + c is the point b b, + c. a 4a If ou forget these formulas, ou can alwas reconstruct them for the specific numeric values of a, b, and c. Copright 00 b Sklight Publishing Eample Find the coordinates of the verte of the parabola Solution. Using the formulas: b 7 7 v = = = ; a 4. If ou forgot the formulas: v = b 49 5 = + c= + 3 =. 4a 4 8 = 7+ 3= + 3 v 7. v = 7 + 3= + 3= + 3= is halfwa between 0 and 7, so 7 v =. 4

27 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 7 If a parabola opens upward and its verte is above the -ais, or the parabola opens downward and its verte is below the -ais, then it does not intersect the -ais so the equation a + b + c = 0 has no solutions in real numbers. This happens b b + 4ac b + 4ac when a > 0 and + c = > 0 or a < 0 and < 0. To have two 4a 4a 4a b 4ac b 4ac zeros, we must have a > 0 and > 0 or a < 0 and < 0. Either wa, 4a 4a we must have d = b 4ac> 0. If d = 0, the verte of the parabola is on the -ais, and the two zeros merge into one. Copright 00 b Sklight Publishing The epression d = b 4ac is called the discriminant of the quadratic function a + b + c. The quadratic equation a + b + c = 0 has two distinct solutions in real numbers if and onl if d > 0. If d < 0, there are no solutions in real numbers. If d = 0, the two solutions merge into one. One wa to derive the formula for the solutions of a quadratic equation (called the quadratic formula) is the technique called completing the square (Figure 6-8). If a quadratic equation a + b + c = 0 has solutions in real numbers, b+ b 4ac b b 4ac these solutions are = and =. a a Notice the discriminant under the square root in the quadratic formula; this confirms our earlier finding that two distinct solutions in real numbers eist if and onl if the discriminant is positive, that the merge into one solution when the discriminant is zero, and that there are no real-number solutions when the discriminant is negative.

28 8 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS + b Also notice that = = v, the -coordinate of the verte of the parabola, as a epected. a + b + c = [factoring out a] b c a + + = a a [adding and subtracting b 4a to complete the square] = a 4a 4a a b b b c a [collecting the first three and the last two terms] a b b 4ac + 0 = a 4a b b 4ac + = a 4a b b 4ac + =± a a ± = b b 4ac a Copright 00 b Sklight Publishing Figure 6-8. Deriving the quadratic formula b completing the square Eample Solve the equation 3 = 0. Solution ± ( ) ± 6 B the quadratic formula, = =. The solutions are = = and = = Eample 3 Solve the equation = 0.

29 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 9 Solution The discriminant of this quadratic function is 5 4 7= 5 8= 3< 0, so there are no real solutions. If f ( ) = a + b+ c= a( )( ), then, clearl, and are the zeros of f. The converse is also true: if and are the zeros of f, then f ( ) can be factored as f ( ) = a( )( ). (This is true even when and are the same number.) When we open the parentheses and collect the like terms, we get a + b + c = a a( + ) + a. This must be true for all, so we must have b c a ( + ) = b + = and a = c =. This fact is known a a as Viète s Theorem (a.k.a. Vieta s Theorem). and are the solutions to the quadratic equation a + b + c, if and b c onl if + = and =. a a Copright 00 b Sklight Publishing This is consistent, of course, with the quadratic formula: if b b 4ac =, then a b b + = = a a and ( b + b 4ac )( b b 4ac ) = = ( a) ( ) ( ) ( a) ( ) b b 4ac b b 4ac 4ac c = = =. 4a 4a a b+ b 4ac = and a Sometimes ou can spot eas factorization for a quadratic function based on Viète s Theorem, especiall when the coefficients are integers. Then ou know the zeros right awa.

30 30 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS Eample 4 Solve the equation 3+ = 0. Solution + = 3 and =, so 3+ = ( )( ). The solutions are = and =. Eercises. Look up the histor of the quadratic formula on the Internet. Roughl when was it discovered?. When and where did François Viète (of Viète s Theorem fame) live? In Question 3-6, find the zeros of a quadratic function (if an) and the coordinates of the verte of its graph. Copright 00 b Sklight Publishing 3. f( ) = ( ) = f f f f ( ) = + + ( ) = ( ) = Factor 5 4. Hint: First find the zeros using the quadratic formula. 9. Can the graphs of two different quadratic functions intersect in more than two points? If es, give an eample; if not, eplain wh not. 0. Write an equation solutions. + p+ q such that = 3 and = 7 are its two

31 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 3. Find the sides of a rectangle whose area is 40 and whose perimeter is 5.. Find the distance between the points A and B at which the line = intersects the parabola =. Then find the area of the region bounded b the parabola and the chord AB. Hint: See Questions <...> and <...> in Section <...>. 3. A line tangent to the parabola = passes through the point (0, ). Find the slope of the line and the point of tangenc. Hint: The slope must be such that the line touches the parabola in eactl one point Factor + + into two quadratic terms. Hint: Set zeros using the quadratic formula. = u and find the 5. Using the quadratic formula, find the Golden Ratio ϕ as the positive solution to the equation =. Using Viète s Theorem, show that the other solution is. ϕ Copright 00 b Sklight Publishing 6. A flare is shot up from a boat over the sea with an initial vertical velocit of gt v = 400 ft/sec. Its vertical position over the water after t seconds is vt, where g 3 ft/sec is the acceleration due to gravit. For how long will the flare fl? What altitude will it reach? 7. Find on YouTube a trailer for the film October Sk. In the film, a group of kids interested in rocketr are accused of starting a forest fire with their model rocket, but Homer Hickam, the main character, learns how to calculate the trajectories of rockets and shows that their rocket couldn t possibl fl that far. (The film is based on a true stor; Homer Hickam eventuall became a NASA engineer.) Suppose a rocket, when launched straight up, reaches an altitude of approimatel 00 meters. If the same rocket is launched at 45 degrees to the surface (the optimal angle for the greatest distance), approimatel how far from the launching place will the rocket land? Hint: The trajector of a rocket is approimatel a parabola; see Question 6.

32 3 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 8. Suppose,, and 3 are solutions to the cubic equation 3 a + b + c + d = 0. Epress and 3in terms of a, b, c, and d. 6.6 Review Concepts, terms, methods, and formulas introduced in this chapter: Copright 00 b Sklight Publishing Linear relation: p + q = C Intercept form of a linear relation: + = a b Parallel lines Perpendicular lines p + q = C is perpendicular to ( q ) + p= C Linear function Slope-intercept form of an equation of a line: = m + b Point-slope form of an equation of a line: = m ( ) = m + b and = + b are perpendicular to each other m Quadratic function: f ( ) = a + b+ c Parabola: Focus Directri Ais Verte Coordinates of the verte: b v =, a v b = + c 4a Quadratic formula for zeros of a quadratic function:, ± = Discriminant: d = b 4ac Viète s Theorem for zeros and of a quadratic function b c + = ; = a a b b 4ac a a + b + c :