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Graph quadratic equations. Complete the square to graph quadratic equations. Use the Vertex Formula to graph quadratic equations. Solve a Quadratic Equation by factoring and square root property. Complete the square to solve quadratics. Use the quadratic formula to solve quadratics. Solve for a Specified Variable Understand the properties of the Discriminant Objectives Students will learn how to;

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Quadratic Function A function is a quadratic function if where a, b, and c are real numbers, with a ≠ 0.

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Simplest Quadratic x (x)(x) – 2– 24 – 1– 11 00 11 24 2 3 2 – 2– 2 3 – 2– 2 4 – 3– 3 – 4– 4 – 3– 3 – 4– 4 4 range [0, ) domain (− , ) x y

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Simplest Quadratic Parabolas are symmetric with respect to a line. The line of symmetry is called the axis of symmetry of the parabola. The point where the axis intersects the parabola is the vertex of the parabola. Vertex Axis Opens up Opens down

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Applying Graphing Techniques to a Quadratic Function The graph of g (x) = ax 2 is a parabola with vertex at the origin that opens up if a is positive and down if a is negative. The width of the graph of g (x) is determined by the magnitude of a. The graph of g (x) is narrower than that of (x) = x 2 if a > 1 and is broader (wider) than that of (x) = x 2 if a < 1. By completing the square, any quadratic function can be written in vertex form the graph of F(x) is the same as the graph of g (x) = ax 2 translated h units horizontally (to the right if h is positive and to the left if h is negative) and translated k units vertically (up if k is positive and down if k is negative).

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Example 1 GRAPHING QUADRATIC FUNCTIONS Solution a. Graph the function. Give the domain and range. x (x)(x) – 1– 13 0– 2– 2 1– 5– 5 2– 6– 6 3– 5– 5 4– 2– 2 53 2 3 – 2– 2 – 6– 6 Domain (− , ) Range [– 6, )

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Example 1 GRAPHING QUADRATIC FUNCTIONS Solution b. Graph the function. Give the domain and range. Domain (− , ) Range (– , 0] 2 3 – 2– 2 – 6– 6

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Example 1 GRAPHING QUADRATIC FUNCTIONS Solution c. Graph the function. Give the domain and range. Domain (− , ) Range (– , 3] (4, 3) 3 – 2– 2 – 6– 6 x = 4

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Example 2 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution Express x 2 – 6x + 7 in the form (x– h) 2 + k by completing the square. Graph by completing the square and locating the vertex. Complete the square.

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Example 2 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution Express x 2 – 6x + 7 in the form (x– h) 2 + k by completing the square. Graph by completing the square and locating the vertex. Add and subtract 9. Regroup terms. Factor; simplify. This form shows that the vertex is (3, – 2)

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Example 2 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution Graph by completing the square and locating the vertex. Find additional ordered pairs that satisfy the equation. Use symmetry about the axis of the parabola to find other ordered pairs. Connect to obtain the graph. Domain is (− , )Range is [–2, )

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Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution To complete the square, the coefficient of x 2 must be 1. Graph by completing the square and locating the vertex. Factor – 3 from the first two terms.

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Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution Graph by completing the square and locating the vertex. Distributive property Be careful here. Factor; simplify.

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Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution Graph by completing the square and locating the vertex. Factor; simplify.

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Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution Intercepts are good additional points to find. Here is the y-intercept. Graph by completing the square and locating the vertex. Let x = 0.

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Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE Solution The x-intercepts are found by setting (x) equal to 0 in the original equation. Graph by completing the square and locating the vertex. Let (x) = 0. Multiply by –1; rewrite. Factor. Zero-factor property

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Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE 2 2 The y-intercept is 1 This x-intercept is – 1. This x-intercept is 1/3. The x intercepts or zeros are the solutions to the quadratic equations

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Graph of a Quadratic Function The quadratic function defined by (x) = ax 2 + bx + c can be written as where Vertex Formula

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Graph of a Quadratic Function The graph of a quadratic function has the following characteristics. Vertex form 1.It is a parabola with vertex (h, k) and the vertical line x = h as axis of symmetry. 2.It opens up if a > 0 and down is a < 0. 3.It is broader than the graph of y = x 2 if a 1. 4.The y-intercept is (0) = c.

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Example 4 FINDING THE AXIS,VERTEX, SOLUTIONS AND GRAPH OF A PARABOLA. Solution Here a = 2, b = 4, and c = 5. The axis of the parabola is the vertical line Find the axis, vertex, solutions and graph of the parabola having equation (x) = 2x 2 +4x + 5 using the vertex formula. The vertex is (– 1, (– 1)). Since (– 1) = 2(– 1) 2 + 4 (– 1) +5 = 3, the vertex is (– 1, 3). Axis is x = -1

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Example 4 FINDING THE AXIS,VERTEX, SOLUTIONS AND GRAPH OF A PARABOLA. XY -311 -25 3 05 111 Axis; x = -1 No real solutions because the graph does not have a value at y = 0; the graph does not cross the x-axis This function has solutions at x=3 and x=6. These values are also called zeros because the y value is zero for x=3 and x=6

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Zero-Factor Property If a and b are numbers with ab = 0, then a = 0 or b = 0 or both.

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Example 1 USING THE ZERO-FACTOR PROPERTY Solve Solution: Standard form Factor. Zero-factor property.

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Example 1 USING THE ZERO-FACTOR PROPERTY Solve Solution: Zero-factor property. Solve each equation.

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Square-Root Property A quadratic equation of the form x 2 = k can also be solved by factoring Subtract k. Factor. Zero-factor property. Solve each equation.

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Square Root Property If x 2 = k, then

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Square-Root Property That is, the solution of If k = 0, then this is sometimes called a double solution. Both solutions are real if k > 0, and both are imaginary if k < 0 If k < 0, we write the solution set as

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Example 2 USING THE SQUARE ROOT PROPERTY a. Solution: By the square root property, the solution set is Solve each quadratic equation.

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Example 2 USING THE SQUARE ROOT PROPERTY b. Solution: Since the solution set of x 2 = − 25 is Solve each quadratic equation.

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Example 2 USING THE SQUARE ROOT PROPERTY c. Solution: Use a generalization of the square root property. Generalized square root property. Add 4. Solve each quadratic equation.

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Solving A Quadratic Equation By Completing The Square To solve ax 2 + bx + c = 0, by completing the square: Step 1 If a ≠ 1, divide both sides of the equation by a. Step 2 Rewrite the equation so that the constant term is alone on one side of the equality symbol. Step 3 Square half the coefficient of x, and add this square to both sides of the equation. Step 4 Factor the resulting trinomial as a perfect square and combine like terms on the other side. Step 5 Use the square root property to complete the solution.

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Example 3 USING THE METHOD OF COMPLETING THE SQUARE a = 1 Solve x 2 – 4x –14 = 0 by completing the square. Solution Step 1 This step is not necessary since a = 1. Step 2 Add 14 to both sides. Step 3 add 4 to both sides. Step 4 Factor; combine terms.

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Example 3 USING THE METHOD OF COMPLETING THE SQUARE a = 1 Solve x 2 – 4x –14 = 0 by completing the square. Solution Step 4 Factor; combine terms. Step 5 Square root property. Take both roots. Add 2. Simplify the radical. The solution set is

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Example 4 USING THE METHOD OF COMPLETING THE SQUARE a ≠ 1 Solve 9x 2 – 12x + 9 = 0 by completing the square. Solution Divide by 9. (Step 1) Add – 1. (Step 2)

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Example 4 USING THE METHOD OF COMPLETING THE SQUARE a = 1 Solve 9x 2 – 12x + 9 = 0 by completing the square. Solution Factor, combine terms. (Step 4) Square root property

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Example 4 USING THE METHOD OF COMPLETING THE SQUARE a = 1 Solve 9x 2 – 12x + 9 = 0 by completing the square. Solution Quotient rule for radicals Square root property Add ⅔.

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Example 4 USING THE METHOD OF COMPLETING THE SQUARE a = 1 The solution set is Solution Add ⅔. Solve 9x 2 – 12x + 9 = 0 by completing the square.

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The Quadratic Formula The method of completing the square can be used to solve any quadratic equation. If we start with the general quadratic equation, ax 2 + bx + c = 0, a ≠ 0, and complete the square to solve this equation for x in terms of the constants a, b, and c, the result is a general formula for solving any quadratic equation. We assume that a > 0.

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Quadratic Formula The solutions of the quadratic equation ax 2 + bx + c = 0, where a ≠ 0, are

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Caution Notice that the fraction bar in the quadratic formula extends under the – b term in the numerator.

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Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Solve x 2 – 4x = – 2 Solution: Write in standard form. Here a = 1, b = – 4, c = 2 Quadratic formula.

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Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Solve x 2 – 4x = – 2 Solution: Quadratic formula. The fraction bar extends under – b.

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Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Solve x 2 – 4x = – 2 Solution: The fraction bar extends under – b.

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Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Solve x 2 – 4x = – 2 Solution: Factor first, then divide. Factor out 2 in the numerator. Lowest terms. The solution set is

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Example 6 USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS) Solve 2x 2 = x – 4. Solution: Write in standard form. Quadratic formula; a = 2, b = – 1, c = 4 Use parentheses and substitute carefully to avoid errors.

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Example 6 USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS) Solve 2x 2 = x – 4. Solution: The solution set is

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Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve for the specified variable. Use when taking square roots. Solution a. Goal: Isolate d, the specified variable. Multiply by 4. Divide by .

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Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution a. See the Note following this example. Square root property Divide by .

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Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution a. Square root property Rationalize the denominator.

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Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution a. Rationalize the denominator. Multiply numerators; multiply denominators.

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Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution a. Multiply numerators; multiply denominators. Simplify.

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Solving for a Specified Variable Note In Example 8, we took both positive and negative square roots. However, if the variable represents a distance or length in an application, we would consider only the positive square root.

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Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution b. Write in standard form. Now use the quadratic formula to find t.

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Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution b. a = r, b = – s, and c = – k

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Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution b. a = r, b = – s, and c = – k Simplify.

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The Discriminant The Discriminant The quantity under the radical in the quadratic formula, b 2 – 4ac, is called the discriminant. Discriminant

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The Discriminant Discriminant Number of Solutions Type of Solution Positive, perfect square TwoRational Positive, but not a perfect square TwoIrrational Zero One (a double solution) Rational NegativeTwo Nonreal complex

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Caution The restriction on a, b, and c is important. For example, for the equation the discriminant is b 2 – 4ac = 5 + 4 = 9, which would indicate two rational solutions if the coefficients were integers. By the quadratic formula, however, the two solutions are irrational numbers,

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Example 9 USING THE DISCRIMINANT Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. Solution a. For 5x 2 + 2x – 4 = 0, a = 5, b = 2, and c = – 4. The discriminant is b 2 – 4ac = 2 2 – 4(5)(– 4) = 84 The discriminant is positive and not a perfect square, so there are two distinct irrational solutions.

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Example 9 USING THE DISCRIMINANT Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. Solution b. First write the equation in standard form as x 2 – 10x + 25 = 0. Thus, a = 1, b = – 10, and c = 25, and b 2 – 4ac =(– 10 ) 2 – 4(1)(25) = 0 There is one distinct rational solution, a “double solution.”

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Example 9 USING THE DISCRIMINANT Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. Solution c. For 2x 2 – x + 1 = 0, a = 2, b = –1, and c = 1, so b 2 – 4ac = (–1) 2 – 4(2)(1) = –7. There are two distinct nonreal complex solutions. (They are complex conjugates.)

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