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QUADRATIC EQUATIONS AND FUNCTIONS
CHAPTER 7 QUADRATIC EQUATIONS AND FUNCTIONS
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7-1 Completing the Square
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Completing the Square Transform the equation so that the constant term c is alone on the right side. If a, the coefficient of the second-degree 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)
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Factor the left side as the square of a binomial.
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
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Solve: a2 – 5a + 3= 0 Move the 3 to the other side a = 1
Complete the square, add (5/2)2 Factor Solve
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Solve: x2 – 6x - 3= 0 Move the 3 to the other side a = 1
Complete the square, add (3)2 Factor Solve
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Solve: 2y2 + 2y + 5 = 0 move the 5 to the other side
divide both sides by 2 (a 1) Add (1/2)2 to both sides Factor Solve
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Solve: 7x2 – 8x + 3 = 0 move the 3 to the other side
divide both sides by 7 (a 1) Add (4/7)2 to both sides Factor Solve
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7-2 Quadratic Formula
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The Quadratic Formula The solutions of the quadratic equation ax2 + bx + c = 0 (a 0) are given by the formula:
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Solve 3x2 + x – 1 = 0 5y2 = 6y – 3 2x2 – 3x + 7 = 0
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7-3 The Discriminant
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Discriminant The discriminant is used to determine the nature of the roots of a quadratic equation and is equal to: D = b2 – 4ac
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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.
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Find the Discriminant x2 + 6x – 2 = 0 3x2 – 4x√3 + 4 = 0 x2 – 6x + 10 = 0
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Discriminant The discriminant also show you whether a quadratic equation with integral coefficients has rational roots. D = b2 – 4ac
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Test for Rational Roots
If a quadratic equation has integral coefficients and its discriminant is a perfect square, then the equation has rational roots.
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Test for Rational Roots
If the quadratic equation can be transformed into an equivalent equation that meets this test, then it has rational roots.
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Find the Determinant and Identify the Nature of the Roots
3x2 - 7x + 5 = 0 2x2 - 13x + 15 = 0 x2 + 6x + 10 = 0
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Equations in Quadratic Form
7-4 Equations in Quadratic Form
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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.
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Example (3x – 2)2 – 5(3x -2) – 6 = 0 Let z = 3x – 2, then z2 – 5z – 6 = 0 Solve for z and then solve for x
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Solve Using Quadratic Form
(x + 2)2 – 5(x + 2) – 14 = 0 (3x + 4)2 + 6(3x + 4) – 16 = 0 x4 + 7x2 – 18 = 0
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7-5 Graphing y – k = a(x- h)2
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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
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Parabola The point where the parabola crosses it axis is the vertex. The graph is a smooth curve.
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Parabola The graph of an equation having the form y – k = a(x - h)2 has a vertex at (h, k) and its axis is the line x = h.
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Graph y = x2 Use the form y – k = a(x – h)2 k= 0, h = 0, so the vertex is (0,0) and x = 0
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Table of Values 1 -1 2 4 -2
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Graph y = ½x2 Use the form y – k = a(x – h)2 k= 0, h = 0, so the vertex is (0,0) and x = 0
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Graph y = -½x2 Use the form y – k = a(x – h)2 k= 0, h = 0, so the vertex is (0,0) and x = 0
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Graph The graph of y = ax2 opens upward if a> 0 and downward if a< 0. The larger the absolute value of a is, the “narrower” the graph.
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Graph y = ½(x-3)2 Use the form y – k = a(x – h)2 k= 0, h = 3, so the vertex is (3,0) and x = 3
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Graph To graph y = a(x – h)2, slide the graph of y = ax2 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.
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Graph y – 3 = ½x2 Use the form y – k = a(x – h)2 k= 3, h = 0, so the vertex is (0,3) and x = 0
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Graph y + 3 = ½x2 Use the form y – k = a(x – h)2 k= -3, h = 0, so the vertex is (0,-3) and x = 0
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Graph To graph y – k = ax2, slide the graph of y = ax2 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.
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7-6 Quadratic Functions
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Completed square form:
Quadratic Functions A function that can be written in either of two forms. General form: f(x) = ax2 + bx + c Completed square form: a(x-h)2 + k
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Graph f(x) = 2(x – 3)2 + 1 y = 2(x – 3)2 + 1
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Graph It’s a parabola with vertex (3,1) and axis x = 3
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Graph f(x) = 3x2 – 6x + 1 y = 3x2 – 6x + 1
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Graph y = 3x2 – 6x + 1 Rewrite the equation in the form : y – k = a(x – h)2 y -1 = 3(x2 – 2x) y – = 3(x2 – 2x + 1) y + 2 = 3(x-1)2
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Quadratic Functions Let f(x) = ax2 + bx + c, a0 If a < 0, f has a maximum value. If a > 0, f has a minimum value and
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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.
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Quadratic Functions Let f(x) = ax2 + bx + c, a0 This maximum or minimum value of f is the y-coordinate when x = - b/2a, at the vertex of the graph.
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Example y = 1/2x2 + 3x – 7/4 Find the maximum or minimum value of f. Find the vertex of the graph of f
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Example y = 1/2x2 + 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
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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)
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Writing Quadratic Equations and Functions
7-7 Writing Quadratic Equations and Functions
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and Theorem A quadratic equation with roots r1 and r2 is
x2 – (r1 + r2)x + r1r2 = 0 or a[x2 – (r1 + r2)x + r1r2 ] = 0 and
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Theorem The equation just given is equivalent to a[x2 – (sum of roots)x + product of roots ] = 0
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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
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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
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Example Find the roots of 2x2 + 9x + 5 = 0 r1 = -9 + 41 r2 = -9 - 41
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Check 4 4 = -18/4 = - b/a r1 · r2 = -9 + 41 · -9 - 41 4 4
= -18/4 = - b/a r1 · r2 = -9 + 41 · -9 - 41 = 5/2 = c/a
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