The xintercepts of the graph are the xvalues for the points where the graph intersects the xaxis. A parabola may have one, two, or no xintercepts.


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1 Chapter 101 Identify Quadratics and their graphs A parabola is the graph of a quadratic function. A quadratic function is a function that can be written in the form, f(x) = ax 2 + bx + c, a 0 or y = ax 2 + bx + c, a 0 This form is the standard form of a quadratic function. When a quadratic function is in the standard form y = ax 2 + bx + c, a 0, the value of the leading coefficient, a, determines the direction that the parabola opens. When a > 0 the parabola opens upward. When a < 0 the parabola opens downward. Positive Quadratic y = x 2 Negative Quadratic y = x 2 The point at which the parabola changes directions is called the vertex of the parabola. If the parabola opens upward the vertex is the lowest point or minimum value of the function. If the parabola opens downward the vertex is the highest point or maximum value of the function. The domain of a quadratic function f(x) = ax 2 + bx + c, a 0, is the set of all real numbers. If the parabola opens upward the range of the function will be all values greater than or equal to the minimum value. If the parabola opens downward the range of the function will be all values less than or equal to the maximum value. You can fold the graph of any quadratic function y = ax 2 + bx + c, a 0 at the vertical line through its vertex, and the two halves match exactly. This fold line is called the axis of symmetry. The vertex of a parabola is the only point on the parabola that is on the axis of symmetry. The xintercepts of the graph are the xvalues for the points where the graph intersects the xaxis. A parabola may have one, two, or no xintercepts.
2 Chapter 102 Graph Quadratic Functions: Parabola You can graph a quadratic function by making a function table. It is helpful to always include the vertex as one of the points in the table when graphing a quadratic function. For the graph of a quadratic function f(x) = ax 2 + bx + c, or y = ax 2 + bx + c, where a, b and c are real numbers and a 0. x = (b/2a) is the equation of its axis of symmetry The xcoordinate of its vertex is (b/2a), to find the yvalue of the vertex substitute the xvalue into the original equation. To graph a quadratic function: Find the equation of the axis of symmetry Find the coordinates of the vertex Make a function table. Select three x values greater than the xcoordinate of the vertex and three x values less than the xcoordinate of the vertex. The more points in your table the more accurate your graph. Example: Graph y = x 2 8x Find the axis of symmetry x = (8/2*1)) = 8/2 = 4 2. Find the coordinates of the vertex. The x coordinate = 4, the ycoordinate is: y = (4) 2 8(4) +12 = = 4 Vertex is (4,4) 3. Make a function table 4. Graph the ordered pairs in the table on the coordinate plane. Draw a smooth curve through them. x y = x 2 8x + 12 (x,y) 1 (1) 2 8(1) +12 = 5 (1,5) 2 (2) 2 8(2) +12 = 0 (2,0) 3 (3) 2 8(3) + 12 = 3 (3,3) 4 (4) 2 8(4) +12 = 4 (4,4) 5 (5) 2 8(5) + 12 = 3 (5,3) 6 (6) 2 8(6) + 12 = 0 (6,0) 7 (7) 2 8(7) + 12 = 5 (7,5)
3 Chapter 10.3 Solving Quadratic Equations by Factoring The equation x 2 +x 20 = 0 is a quadratic equation in one variable that is in standard form. The equation ax 2 + bx + c = 0 where a, b and c are real numbers and a 0 is a quadratic equation that is written in standard form. The values of the variable in an equation that make the equation true are called solutions of the equation. We can solve a quadratic equation by factoring. To do this we must apply the Zero Product Property which states that if ab = 0 then either a = 0 or b = 0, where a and b are real numbers. This property enables you to conclude that if a product is equal to 0, then at least one of its factors must be equal to 0. In our example x 2 +x 20 = 0 if we factor the quadratic we get (x+5)(x4) = 0. The Zero Product Property states that one of the factors must equal 0. So either (x+5) = 0 or (x4) = 0. We can solve these equations to determine that either x = 5 or x = 4. So the solution set for x 2 +x 20 = 0 is {5,4}. The solutions of the quadratic equation, ax 2 + bx + c = 0, are the xintercepts of the related quadratic function, f(x) = ax 2 + bx + c. The solutions of the equation are called the roots of the equation or zeros of the function. Sometimes there is only one solution to a quadratic equation. In this case the xintercept is also the vertex of the quadratic function. Not every quadratic equation can be factored over integers. Sometimes the roots of a quadratic equation are irrational numbers, or the roots are not real numbers. The xintercepts of quadratics that cannot be factored can be approximated through graphing. If a quadratic does not have xintercepts then it does not have real roots. Some equations containing radicals result in quadratic equations. Sometimes the quadratic equations will have extra roots, called extraneous roots, so not all the roots of the quadratic equations are solutions to the original equations. Solutions to quadratic equations are given in set notation. One important issue should be mentioned at this point: Just as with linear equations, the solutions to quadratic equations may be verified by plugging them back into the original equation, and making sure that they work, that they result in a true statement. For the above example, we would do the following: Checking x = 3 in (x 3)(x 4) = 0: Checking x = 4 in (x 3)(x 4) = 0: ([3] 3)([3] 4)?=? 0 ([4] 3)([4] 4)?=? 0 (3 3)(3 4)?=? 0 (4 3)(4 4)?=? 0 (0)( 1)?=? 0 (1)(0)?=? 0 0 = 0 0 = 0
4 Chapter 10.4 Solve Verbal Problems Involving the Quadratic Equations Mrs. Baca's art class is painting a mural on the front of the school. The mural is 3 meters wider than it is high and has an area of 10m 2. What are the dimensions of the mural. To solve this problem you may want to begin by drawing a sketch of the mural. We can let x = the height of the of the mural, so the width will be x+3. The area is 10m 2. So we can set up the equation 10m 2 = x(x+3) If we multiply this we get 10 = x 2 + 3x. We then want to write the equation in standard form by subtracting 10 from each side of the equation. x 2 + 3x 10 = 0 Now we want to factor the quadratic (x+5)(x2) = 0 Now we want to apply the Zero Product Property x+5 = 0 or x2 = 0 x = 5 or x = 2 You know that since x represents the height of the mural so x = 5 can be eliminated as a solution for x since you cannot have a negative height. The height of the mural will be 2 meters and the width will be 5 meters. You can write a quadratic equation, given its roots, by working backward. From the roots write the binomial factors Multiply the factors to write the equation in standard form. Example: Write an equation with given roots {3,5} If the roots are 3 and 5 then the factors are (x+3) and (x5) (x+3)(x5) = 0 x 2 +3x5x15 = 0 x 2 2x 15 = 0 is an equation with roots {3,5}
5 Chapter 10.5 Solve Quadratic Equations by Completing the Square To solve a quadratic equation by completing the square you must make the quadratic expression on one side of the equation into a perfect square. After completing the square you can solve the equation by taking the square root of each side. Completing the square is often a good method to use for solving a quadratic equation when the equation is not factorable using intergers. Completing the Square to solve ax 2 + bx + c = 0 where a, b and c are real numbers and a 0: Write the equation so that the constant term, c, is isolated on the right side. Divide each side of the equation by a; x 2 +(b/a)x = c/a Find the square of one half the coefficient of x. Add that number to each side of the equation. Factor the left side of the equation. The results should have the form (x + r) 2 where r is a constant. Take the square root of each side. Then solve for x, and simplify the solutions. Example: Solve 3x 2 7x +2 = 0 3x 2 7x = 2 x 2 7/3x = 2/3 ½ of 7/3 = 7/6 x 2 7/3x + 49/36 = 2/3 + 49/36 (x7/6) 2 = 24/ /36 (x7/6) 2 = 25/36 (x7/6) 2 = 25/36 rewrite equation with constant isolated on right side of equation divide both sides of the equation by a Find ½ the value of the x term add the squared value of ½ x to both sides of the equation. make like denominators and add the values on the side of equation rewrite the trinomial as the square of a binomial take the square root of both sides of the equation x7/6 = 5/6 or x 7/6 = 5/6 write and solve two equations. x7/6 + 7/6 = 5/6 +7/6 x 7/6 + 7/6 = 5/6 + 7/6 x = 12/6 = 2 x = 2/6 = 1/3 So, x = 2 or x = 1/3 It is important to remember that every positive real number has two square roots one positive and one negative.
6 Chapter 10.6 The Quadratic Formula and the Discriminant The Quadratic Formula is a formula that can be used to solve any quadratic equation. It is derived by solving the standard form of a quadratic equation, ax 2 + bx + c = 0, for x by completing the square. The Quadratic Formula is as follows: If ax 2 + bx + c = 0, where a 0, then x = b +/ b 24ac 2a The expression b 24ac is called the discriminant of the quadratic equation, and it provides important information about the roots of the equation. The symbol +/ indicates that the discriminant will be both added and subtracted, so the equation will have two roots. If b 24ac = 0 then the only root of the equation is b/2a, which is a rational number. If b 24ac > 0 then b +/ positive number. If b 24ac is a perfect square, both roots will be 2a rational. If it is not, the two roots will be irrational. If b 24ac < 0 then x = b +/ negative number. The equation will have no real roots because 2a the square root of a negative number is not a real number.
7 Chapter 10.7 Solve Quadratic Equations with the Quadratic Formula The equation 3x x 275 can be solved by factoring or completing the square, however a more practical method would be to use the Quadratic Formula. Remember that the Quadratic Formula is: If ax 2 + bx + c = 0, where a 0, then x = b +/ b 24ac 2a In our example: a = 3, b = 40 and c = 275 x = 40 +/ (40) 2 4(3)(275) 2(3) x = 40 +/ (3300) 6 x = 40 +/ x = 40 +/ x = = 30 or x = = x = 5 or x = 55/3
8 Chapter 10.8 Solve LinearQuadratic Systems One way to solve a system of linearquadratic equations is by substitution. Solve the linear equation for y and substitute that value into the quadratic equation. Let s try it: y = x 2 + 5x + 18 y = 2x + 18 y = 2x + 18 is already solved for y so all we have to do is substitute the value for y in the linear equation into the quadratic equation. 2x + 18 = x 2 + 5x x +2x addition property of equality 18 = x 2 + 7x subtraction property of equality 0 = x 2 + 7x 0 = x(x+7) distributive property x = 0 or x+7 = 0 zero product property x = 0 or x = 7 Now to find y we have to substitute the values for x into either equation. If x = 0 then y = 2(0) + 18, y = 18 If x = 7, then y = 2(7) + 18, y = , y = 32 (0,18) is a solution and (7,32) is a solution Systems of linearquadratic equations can have two solutions (two points of intersection), one solution (the parabola and line intersect at one point), or no solutions (no intersection). When you solve a linearquadratic system you must check your solution in both equations in the system. What you are doing when you solve a system is finding the point or points where the two graphs intersect. That point or points when substituted into the original equations will make them both true. If the point does not make both equations true it is not a solution to the system. When solving a linearquadratic system of equations: Solve one equation for one variable in terms of the other Substitute into the equation not used in step 1 Solve the equation Substitute to find the corresponding values of the other variable Check the solutions
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