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CHAPTER 7 QUADRATIC EQUATIONS AND FUNCTIONS 7-1 Completing the Square Completing the Square Transform the equation so that the constant term c is alone on the right side. If a, the coefficient of the seconddegree term, is not equal to 1, then divide both sides by a. Add the square of half the coefficient of the first-degree term, (b/2a)2, to both sides (Completing the square) Completing the Square Factor the left side as the square of a binomial. Complete the solution using the fact that (x + q)2 = r is equivalent to x + q = ±r Solve: 2 a – 5a + 3= 0 1. Move the 3 to the other side 2. a = 1 3. Complete the square, add (5/2)2 4. Factor 5. Solve Solve: 2 x – 6x - 3= 0 1. Move the 3 to the other side 2. a = 1 3. Complete the square, add (3)2 4. Factor 5. Solve Solve: 2 2y + 2y + 5 = 0 1. move the 5 to the other side 2. divide both sides by 2 (a 1) 3. Add (1/2)2 to both sides 4. Factor 5. Solve Solve: 2 7x – 8x + 3 = 0 1. move the 3 to the other side 2. divide both sides by 7 (a 1) 3. Add (4/7)2 to both sides 4. Factor 5. Solve 7-2 Quadratic Formula The Quadratic Formula The solutions of the quadratic equation 2 ax + bx + c = 0 (a 0) are given by the formula: Solve 2 • 3x + x – 1 = 0 • 5y2 = 6y – 3 2 • 2x – 3x + 7 = 0 7-3 The Discriminant Discriminant The discriminant is used to determine the nature of the roots of a quadratic equation and is equal to: D= 2 b – 4ac Discriminant Cases If D is positive, then the roots are real and unequal. If D is zero, then the roots are real and equal (double root) . negative, the roots are imaginary. Find the Discriminant 2 • x + 6x – 2 = 0 • 3x2 – 4x√3 + 4 = 0 2 •x – 6x + 10 = 0 Discriminant The discriminant also show you whether a quadratic equation with integral coefficients has rational roots. 2 D = b – 4ac Test for Rational Roots If a quadratic equation has integral coefficients and its discriminant is a perfect square, then the equation has rational roots. Test for Rational Roots If the quadratic equation can be transformed into an equivalent equation that meets this test, then it has rational roots. Find the Determinant and Identify the Nature of the Roots • 3x2 - 7x + 5 = 0 2 • 2x - 13x + 15 = 0 • x2 + 6x + 10 = 0 7-4 Equations in Quadratic Form Quadratic Form An equation in quadratic form can be written as: a[f(x)]2 + b[(f(x)] + c = 0 where a 0 and f(x) is some function of x. It is helpful to replace f(x) with a single variable. Example (3x – 2)2 – 5(3x -2) – 6 = 0 Let z = 3x – 2, then 2 z – 5z – 6 = 0 Solve for z and then solve for x Solve Using Quadratic Form (x + 2)2 – 5(x + 2) – 14 = 0 (3x + 4 x + 2 4) 2 7x + 6(3x + 4) – 16 = 0 – 18 = 0 7-5 Graphing y – k = a(x- 2 h) Parabola Parabola is the set of all points in the plane equidistant from a given line and a given point not on the line. Parabolas have an axis of symmetry (mirror image) either the x-axis or the y-axis. and Parabola The point where the parabola crosses it axis is the vertex. The graph is a smooth curve. Parabola The graph of an equation having the form 2 y – k = a(x - h) has a vertex at (h, k) and its axis is the line x = h. Graph y = 2 x Use the form y – k = a(x – 2 h) k= 0, h = 0, so the vertex is (0,0) and x = 0 Table of Values 0 0 1 1 -1 1 2 4 -2 4 qx = x2 4 2 -5 5 -2 -4 Graph y = 2 ½x Use the form y – k = a(x – 2 h) k= 0, h = 0, so the vertex is (0,0) and x = 0 fx = 1 2 x2 gx = x2 4 2 -5 5 -2 -4 Graph y = 2 -½x Use the form y – k = a(x – 2 h) k= 0, h = 0, so the vertex is (0,0) and x = 0 qx = x2 rx = -1 2 x2 4 2 -5 5 -2 -4 Graph 2 ax The graph of y = opens upward if a> 0 and downward if a< 0. The larger the absolute value of a is, the “narrower” the graph. qx = x2 s x = 5x2 4 2 -5 5 -2 -4 Graph y = 2 ½(x-3) Use the form y – k = a(x – 2 h) k= 0, h = 3, so the vertex is (3,0) and x = 3 fx = 1 2 x-3 2 4 2 -5 5 -2 -4 fx = gx = 1 2 1 2 x-3 2 x+32 4 2 -5 5 -2 -4 Graph To graph y = a(x – h)2, slide 2 the graph of y = ax horizontally h units. If h > 0, slide it to the right; if h < 0, slide it to the left. The graph has vertex (h, 0) and its axis is the line x = h. Graph y – 3 = 2 ½x Use the form y – k = a(x – 2 h) k= 3, h = 0, so the vertex is (0,3) and x = 0 12 10 8 6 fx = 1 2 x2+3 4 2 Graph y + 3 = 2 ½x Use the form y – k = a(x – 2 h) k= -3, h = 0, so the vertex is (0,-3) and x = 0 8 6 fx = gx = 1 2 1 2 x2+3 x2-3 4 2 -5 5 -2 Graph To graph y – k = ax2, slide 2 the graph of y = ax vertically k units. If k > 0, slide it upward; if k < 0, slide it downward. The graph has vertex (0, k) and its axis is the line x = 0. 7-6 Quadratic Functions Quadratic Functions A function that can be written in either of two forms. General form: 2 f(x) = ax + bx + c Completed square form: a(x-h)2 + k Graph f(x) = 2(x – y = 2(x – 2 3) 2 3) +1 +1 Graph It’s a parabola with vertex (3,1) and axis x = 3 8 6 4 fx = 2x-32+1 2 -5 5 -2 10 Graph f(x) = y= 2 3x 2 3x – 6x + 1 – 6x + 1 Graph 2 3x y = – 6x + 1 Rewrite the equation in the 2 form : y – k = a(x – h) 2 y -1 = 3(x – 2x) 2 y – 1 + 3 = 3(x – 2x + 1) y + 2 = 3(x-1)2 fx = 3x2-6x+1 6 4 2 -5 5 -2 -4 -6 Quadratic Functions 2 ax Let f(x) = + bx + c, a0 If a < 0, f has a maximum value. If a > 0, f has a minimum value. and Quadratic Functions The graph of f is a parabola. This maximum or minimum value of f is the y-coordinate when x = - b/2a, at the vertex of the graph. Quadratic Functions 2 ax Let f(x) = + bx + c, a0 This maximum or minimum value of f is the y-coordinate when x = - b/2a, at the vertex of the graph. Example 2 1/2x y= + 3x – 7/4 Find the maximum or minimum value of f. Find the vertex of the graph of f Example 2 1/2x y= + 3x – 7/4 a = ½, a > 0, then f has a minimum value. Minimum occurs when x = -b/2a x = -3/2*1/2 = -3 Example y = 1/2x2 + 3x – 7/4 y = ½(-3)2 + 3(-3) – 7/4 y = -25/4 So the minimum value of f is -25/4 and the vertex is (-3, -25/4) 7-7 Writing Quadratic Equations and Functions Theorem A quadratic equation with roots r1 and r2 is x2 – (r1 + r2)x + r1r2 = 0 or 2 a[x – (r1 + r2)x + r1r2 ] = 0 and Theorem The equation just given is equivalent to 2 a[x – (sum of roots)x + product of roots ] = 0 Example Find a quadratic equation with roots (2 + i)/3 and (2 – i)/3 Sum of roots = (2 + i)/3 + (2 – i)/3 = 4/3 Product of roots = (2 + i)/3 (2 – i)/3= 5/9 x2 – 4/3x + 5/9 = 9x2 -12x + 5 Theorem If r1 and r2 are the roots of a quadratic equation ax2 + bx + c =0, then r1 + r2 = sum of roots = -b/a and r1r2 = product of roots = c/a Example Find the roots of 2x2 + 9x + 5 = 0 r1 = -9 + 41 r2 = -9 - 41 4 4 Check r1 + r2 = -9 + 41 + -9 - 41 4 4 = -18/4 = - b/a r1 · r2 = -9 + 41 · -9 - 41 = 5/2 = c/a 4 4 END END