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Quadratic Formula and Equations Quadratic Formula and Equations

A quadratic equation is an equation of the second degree, meaning that for an equation in x, the greatest exponent on x is 2. Quadratics most commonly refer to vertically oriented parabolas—that is, parabolas that open upward or downward. The graph of a vertically oriented parabola has the shape of a rounded "v," and the bottom-most (or top-most) point is called the vertex. The equation for a parabola is usually written in either standard or vertex form; however, the standard form is more commonly used to solve for the x -intercepts, or roots. The standard form is y = ax 2 + bx + c for any real numbers a, b, c where a ≠ 0. The vertex form is y − k = a (x − b )2 with vertex (b, k ) and where a ≠ 0. Because x -intercepts are the points at which the graph crosses the x -axis, the solutions are always found by substituting 0 for y. The roots are often useful in solving real world problems, and there are three common ways to find the roots: factoring, using the quadratic formula, and completing the square. Not all quadratics can easily be factored, but if they can, the quickest way to solve them is to factor and use the zero product property. The zero product property basically states that if the product of two numbers is 0, then at least one of the numbers multiplied must be a 0. In other words, for any real numbers a and b, if ab = 0, then either a = 0 or b = 0. Consider a swimmer who starts at one end of a pool, swims down to pick up a ring at the bottom of the middle of the pool, and then surfaces at the other end of the pool with the ring. The equation y = x 2 − 6x + 9 can be used to model the path of the swimmer, where y is the water level in the pool measured in feet, and x is the time in seconds since the swimmer started. The equations below show how to solve for the roots of the equation to find the number of seconds it took the swimmer to reach the ring at the bottom of the pool: namely, by substituting 0 for y, factoring, and using the zero product property. 0 = x 2 − 6x + 9 Substitute 0 for y. 0 = [x − 3](x − 3) Factor x 2 − 6x + 9. Either [x − 3] = 0 or (x − 3) = 0 Use the zero product property. So x = 3 and x = 3 Solve each equation. In this example, the quadratic has only one repeated root, x = 3. This root is the time at which the swimmer reached the bottom of the pool. This quadratic can be graphed by substituting values into the equation to make a table of points, then graphing the realistic portion of the parabola, as shown below. The graph below illustrates that the parabola has a vertical line of symmetry that passes through (3, 0). The equation for the line of symmetry of this parabola is x = 3. The graph of the parabola continues infinitely; however, to model the path of the swimmer, only the points from 0 to 6 seconds are graphed. This is because the swimmer starts at x = 0 seconds, swims down for 3 seconds to get the ring, and then swims up for 3 seconds. Because not all quadratic equations can be factored, other methods for finding roots are needed. One other method of finding roots for a quadratic is to use the quadratic formula. In the formula, the plus or minus sign means to solve the formula twice—once with a plus, and once with a minus. In other words, given a quadratic equation in standard form, y = ax 2 + bx + c, the solutions can be found by Consider that a delayed space shuttle leaves Earth about 20 minutes after the scheduled departure. At 6 miles out, the shuttle turns around and returns to Earth. The distance of the shuttle from Earth can be described by the equation y − 6 = −0.1(x − 30)2, where x is the number of minutes the shuttle is in flight. The equations below show how to find the total number of minutes the shuttle was off the ground. To find the roots of the equation, first solve for standard form and substitute 0 for y, as shown in Step One. The resulting trinomial cannot easily be factored into two binomials, so the quadratic equation must be used to solve for the roots, as shown in Step Two. Step One Step Two To graph the parabola, plot and connect the two roots and the vertex. (The equation was originally given in vertex form.) If needed, more points can be found by substituting values for x into the equation. To graph the realistic portion of the parabola, graph only the portion in Quadrant I (see below). The original problem said that the shuttle was delayed by about 20 minutes. This is the first intercept, x ≈ 22.25 minutes. The vertex represents the point at which the shuttle was 6 miles from Earth. The second intercept, x ≈ 37.75 represents the time at which the shuttle returned to Earth. To find the total number of minutes the shuttle was in flight, subtract its liftoff and landing times. The shuttle was in flight for about 37.75 − 22.25 = 15.5 minutes. Some equations for parabolas may be solved more easily by completing the square. Completing the square forces the trinomial to be a perfect square by replacing the constant term with . In addition to finding roots, completing the square is also used for transforming an equation in standard form to vertex form. Furthermore, this method can be extended for use with the other conic sections . The quadratic formula can be derived by completing the square in y = ax 2 + bx + c. The equations below show how to solve for vertex form of y = x 2 – 6x + 7 and find its roots by completing the square. The strategy is to move the constant term opposite the trinomial and replace it with Then the new trinomial is written as the square of a binomial. In Step Two, the vertex form is y + 2 = (x − 3)2, and the vertex is (3 − 2). To find the roots, substitute 0 for y and solve for x. The two roots are and . To graph this parabola, the vertex and approximations for the roots can be plotted and connected. Step One Step Two b)2. Then The vertex can be found directly from vertex form, and it can also be found from standard form. From standard form, use b /2a to find the x -coordinate of the vertex and then substitute the result into the equation to find the y -coordinate of the vertex. The discriminant can be used to determine if the graph crosses the x -axis, and if so how many times. The discriminant is the expression under the radical in the quadratic formula, b 2 − 4ac. A square root usually yields two solutions, unless it is the square root of zero. The table summarizes the number and types of solutions that can occur and how they affect the appearance of the graph. The value of a in a quadratic equation also affects the placement of the graph on the plane. If a is positive, the graph opens upwards; if it is negative the graph opens downward. If a is greater than one, the graph will be narrow, and if a is a fraction between 0 and 1, the graph will be wide. This bit of information is especially useful because the value of a affects other types of graphs in the same ways as it does parabolas. All conic sections are quadratics because they have equations of the second degree. However, only the vertically oriented parabolas that have been summarized in this article are functions. Graphing calculators and computers perform functions by taking an input and giving an output. Hence, most graphing tools are only equipped to graph equations of functions. To graph a horizontally oriented parabola on a calculator, the graph must be broken into pieces that are functions. Then the equations for each piece are graphed on the same plane to create the appearance of one graph. see also Conic Sections; Functions and Equations; Graphs and Effects of Parameter Changes. Michelle R. Michael


From Yahoo Answers

Question:The vertex and the axis of symmetry are related, but I am not sure how to explain it. Can someone help me. I know that you find them by using -b-4ac/2a, but that doesn't explain how they are related?

Answers:person above made mistake when calculating dy/dx axis of symmetry is the line x=-b/2a the vertex will ALWAYS lie on the axis of symmetry of a parabola and for his last point, axis of symmetry is x=a, NOT Y

Question:So I'm learning about quadratic equations and parabolas... and I'm just wondering, when I solve for my zeros, and I only get one zero (x-int.) the zero in this case (when there's only one of them) is the equation of the axis of symmetry, right? For example, I solve for my zeros and there is only one. It is 2. (x=2) The equation of the axis of symmetry is also x=2 and the vertex is also located at (2,0) right?

Answers:I suppose your right because for there to be one solution then the parabola would have to have its lowest point, or minimum, in direct contact to a point on the x axis. This point would be the vertex also the area of the line of symmetry

Question:Question 1. What is the domain and range of the quadratic equation y = x2 14x 52? a. Domain: All Real Numbers Range: y 7 b. Domain: All Real Numbers Range: y 7 c. Domain: All Real Numbers Range: y 3 d. Domain: All Real Numbers Range: y 3 Question 2. What are the x and y intercepts of the quadratic equation y = (x 3)2 25? Identify the x intercepts by graphing the quadratic equation. a. x intercepts: ( 2, 0) and (8, 0) y intercept: (0, 16) b. x intercepts: (2, 0) and ( 8, 0) y intercept: (0, 16) c. x intercepts: ( 2, 0) and (8, 0) y intercept: (0, 25) d. x intercepts: (2, 0) and ( 8, 0) y intercept: (0, 25) Question 3. Which of the following is the vertex of the quadratic equation y = 6x2 24x + 32? a. ( 2, 8) b. ( 2, 8) c. (2, 8) d. (2, 8) Question 4. What are the x and y intercepts of the quadratic equation y = 2x2 8x + 6? a. x intercepts: ( 3, 0) and ( 1, 0) y intercept: (0, 3) b. x intercepts: ( 3, 0) and ( 1, 0) y intercept: (0, 6) c. x intercepts: ( 3, 0) and ( 1, 0) y intercept: (0, 3) d. x intercepts: (3, 0) and (1, 0) y intercept: (0, 6) Question 5. Which of the following is the equation of the axis of symmetry of the quadratic equation y = 2x2 + 24x + 62? a. x = 6 b. x = 6 c. x = 10 d. x = 10 Question 6. What is the domain and range of the quadratic equation y = (x 5)2 + 10? a. Domain: All Real Numbers Range: y 5 b. Domain: All Real Numbers Range: y 5 c. Domain: All Real Numbers Range: y 10 d. Domain: All Real Numbers Range: y 10 Question 7. Which of the following is the equation of the axis of symmetry of the quadratic equation y = 2(x + 7)2 4? a. x = 7 b. x = 4 c. x = 4 d. x = 7 Question 8 Which of the following is the vertex of the quadratic equation y = 4(x + 6)2 + 2? a. (6, 2) b. ( 6, 2) c. (6, 2) d. ( 6, 2)

Answers:1) Domain is all real numbers y = x^2 14x 52 y = -(x^2 + 14x + 52) y = -(x^2 + 14x + 49) - 3 y = -(x + 7)^2 - 3 Range is all x <= -3. In set notation, the range is (-infinity, -3].

Question:Hey, If your axis of symmetry is 0, is your vertex automatically 0? thanks

Answers:The x value is automatically 0, but the y value needs to be solved for. In the equation ax^2+c, c will be your y value. Thus (0,c) is the vertex

From Youtube

Moya Math Agebra 2 (Identifying the Vertex and Axis of Symmetry From a Graph) - with Sound.MOV :How to identify the vertex and axis of symmetry when given a quadratic function graph. This video also shows you how to write an equation for the axis of symmetry.

quadratic equations 2b :Graphing a parabola from an equation in standard form. Includes x-intercept, y-intercept, vertex, and axis of symmetry. Be sure to use at least 5 points!