Forms of Quadratic Functions
Quadratic functions can be represented in different forms, including standard form (y = ax^2 + bx + c), vertex form (y = a(x - h)^2 + k), and factored form (y = a(x - r)(x - s)). Each form provides unique insights into the function's properties, such as its vertex, roots, and axis of symmetry. These forms are useful for graphing, solving equations, and understanding quadratic relationships.
5 Key excerpts on "Forms of Quadratic Functions"
eBook - ePubDifferentiating Instruction in Algebra 1
Ready-to-Use Activities for All Students (Grades 7-10)
- Kelli Jurek(Author)
- 2021(Publication Date)
- Routledge(Publisher)
...Solve the problems and show your work on a separate sheet of paper. Circle the problems you choose on this page and then reference them on your work paper with the number from the square. Name:────────── Hour/Block:───── Date:───── Lesson 2: Characteristics of Quadratics Forms of Quadratic Equations General/Standard Form Vertex Form Factored Form y = ax + bx 2 + c, where c is the starting value (y -intercept) of the function y = a (x - h) 2 + k where (h, k) is the vertex of the parabola y = a (x - r 1)(x - r 2), where r 1 and r 2 are the zeros of the quadratic function Directions : Identify the form for each of the following quadratic equations. If the function is written in general/standard form, identify the starting value. For all equations, find the vertex (as a coordinate pair), write the equation for the axis of symmetry (in the form x =), and then find the zeros. There is one sample problem. Round decimals to two places if necessary. Equation Form Starting Value Vertex Axis of Symmetry Zeros y = 2(x + 2) 2 - 4 vertex form — (–2,–4) x = –2 –3.41,–0.59 y = 2(x -2) +5 y = 3(x -2)(x -4) y = -6(x -3) +6 y = 2 x 2 + 4 x - 6 y = x 2 - 3 x + 7 When you have completed the problems, check your answers. Each box is worth 1 point. My score is ───────. If you scored 0-17 points: You need additional practice and will complete the hexagon puzzle labeled with a moon. If you scored 18-22 points: You are working on level but need a little more practice. You will complete the hexagon puzzle labeled with a sun. If you scored 23-25 points: You are ready to be challenged and will complete the hexagon puzzle labeled with a star. Name:────────── Hour/Block:───── Date:───── Lesson 2: Characteristics of Quadratics Hexagon Puzzle Directions : Solve any problem in a hexagon in the far left-hand column. If you get the answer correct, choose a problem from the next column that is adjacent to the first problem...
- Lawrence S. Leff, Christina Pawlowski-Polanish(Authors)
- 2021(Publication Date)
- Barrons Educational Services(Publisher)
...Any linear equation can be solved by writing it in the form ax + b = c, and then substituting the values for a, b, and c in the formula. For example, if 4 x − 3 = 25, then a = 4, b = −3, and c = 25, so Similarly, if ax 2 + bx + c = 0, then x can be solved for in terms of the coefficients a, b, and c. The result is called the quadratic formula. USING THE QUADRATIC FORMULA The method of completing the square can be used to obtain a general solution to the quadratic equation ax 2 + bx + c = 0, in which the roots of the equation are expressed in terms of the coefficients a, b, and c. QUADRATIC FORMULA If ax 2 + bx + c = 0 and a ≠ 0, then Any quadratic equation, including equations with rational, irrational, or imaginary roots, can be solved by writing the equation in the form ax 2 + bx + c = 0, and then substituting the numerical values of a, b, and c into the quadratic formula. EXERCISE 1 Solving a Quadratic Equation by Formula Find the roots of 2 x 2 − 3 x − 1 = 0 in radical form. SOLUTION Use the quadratic formula, where a = 2, b = −3, and c =. −1: If the two roots of a quadratic equation are denoted by x 1 and x 2, then Although you could use a graphing calculator to estimate the roots of this equation, the quadratic formula allows you to obtain an exact representation of the irrational roots in radical form. FINDING NON-REAL ROOTS OF QUADRATIC EQUATIONS If a quadratic equation has non-real roots, its graph does not intersect the x -axis. The quadratic formula can be used to find these non-real roots. EXERCISE 2 Finding the Non-real Roots of a Quadratic Equation Find the roots of x 2 + 5 = 2 x in a + bi form. SOLUTION If x 2 + 5 = 2 x, then x 2 − 2 x + 5 = 0. Use the quadratic formula, where a = 1, b = −2, and c = 5: Hence, x 1 = 1 + 2 i and x 2 = 1 – 2 i. MAKING GRAPHICAL CONNECTIONS The expression b 2 − 4 ac underneath the radical sign in the quadratic formula is called the discriminant...
eBook - ePub- Stu Schwartz(Author)
- 2013(Publication Date)
- Research & Education Association(Publisher)
...Since the roots of the quadratic equation indicate where the quadratic function equals zero, our technique for solving quadratic equations is to set the function equal to zero by moving all the terms to one side of the equal sign, leaving 0 on the other side. Then, to simplify the function, factor it, set each factor equal to zero, and solve for x. Alternatively, the solutions for x can be found by using the quadratic formula, introduced in the next section, especially useful when factoring isn’t straightforward. The values of x found in this way are called solutions of the equation. Graphically, these are the points where the graph of the function crosses or touches the x -axis (where y or f (x) = 0). EXAMPLE Show how the graphs of the following quadratic functions match the solutions to the quadratic equations. a. f (x) = x 2 b. f (x) = x 2 − 4 c. f (x) = x 2 + 1 SOLUTION a. The graph of this common function, y = x 2, as shown in Chapter 4, is the following: This function is called a parabola. It has one root, as seen in the graph. When we set the function equal to zero, we get the equation x 2 = 0, which has one solution, x = 0, as can be seen on the graph. b. As discussed in Chapter 4, y = x 2 − 4 is the function y = x 2 shifted down 4 units. The graph is as follows: The graph shows that the function f (x) = x 2 − 4 has two roots, so the equation y = x 2 − 4 = 0 has two solutions, which can be found by factoring. x 2 − 4 = 0 (x + 2) (x − 2) = 0 x + 2 = 0 x − 2 = 0 x = −2 or x = 2 Note that when there are two solutions, both are true and we can connect them with “or.” c. As also discussed in Chapter 4, y = x 2 + 1 is the function y = x 2 shifted up 1 unit. The graph is as follows: The graph shows that the function f (x) = x 2 + + has no roots since it never touches the x -axis (y ≠ 0). So the equation y = x 2 + 1 has no solutions. This goes hand-in-hand with the fact that x 2 + 1 does not factor...
eBook - ePub- Michael Harrison, Patrick Waldron(Authors)
- 2011(Publication Date)
- Routledge(Publisher)
...Conic sections, quadratic forms and definite matrices DOI: 10.4324/9780203829998-6 f 4.1 Introduction So far, we have concentrated on linear equations, which represent lines in the plane, planes in three-dimensional space or, as will be seen in Section 7.4.1, hyperplanes in higher dimensions. In this chapter, we consider equations that also include second-order or squared terms, and that represent the simplest types of nonlinear curves and surfaces. The concepts of matrix quadratic form and definite matrix are of considerable importance in economics and finance, as will be seen in the detailed study of our several applied examples later. Quadratic forms relate importantly to the algebraic representation of conic sections. Also, in Theorem 10.2.5, it will be seen that the definiteness of a matrix is an essential idea in the theory of convex functions. This chapter gives definitions and simple illustrations of the concepts of quadratic form and definiteness before going on to establish a number of theorems relating to them. We will return to quadratic forms in Chapter 14, where the important general problem of maximization or minimization of a quadratic form subject to linear inequality constraints is studied. 4.2 Conic sections In this section, we consider equations representing conic sections in two dimensions. There are a number of equivalent ways of describing and classifying conic sections. We begin with a geometric approach. 1 Consider the curve traced out in the coordinate plane ℝ 2 by a point P = (x, y), which moves so that its distance from a fixed point (the focus S) is always in a constant ratio (the eccentricity ϵ ≥ 0) to its perpendicular distance from a fixed straight line (the directrix L). This curve is called: an ellipse when 0 < ϵ < 1; a parabola when ϵ = 1; a hyperbola when ϵ > 1; and a circle as ϵ → 0, as we shall see later. 4.2.1 Parabola Consider first the case of ϵ = 1, i.e. the parabola...
- F. Xavier Malcata(Author)
- 2020(Publication Date)
- Wiley(Publisher)
...(7.83). The functional form of Eq. (7.109) matches, as expected, that of Eq. (2.182) – after replacing a 2 by a and r k by x k (k = 1, 2). A graphical solution of the quadratic equation was developed in 1867 by Austrian engineer Eduard Lill – and is outlined in Fig. 7.1. If the three germane coefficients a, b, and c represent the lengths of straight segments [ AB ], [ BC ], and [ CD ], respectively, i.e.,, and, holding right angles between them – then the circle with diameter [ AD ] crosses [ BC ] in up to two points, E and F ; the solutions | x 1 | and | x 2 | of Eq. (7.83) are then given by / and /. Figure 7.1 Graphical solution of quadratic equation ax 2 + bx + c = 0, departing from straight segments [ AB ] with length a, [ BC ] with length b, and [ CD ] with length c – all at right angles with each other; and intersection of the circle of diameter [ AD ] with straight segment [ BC ] at points E and F, corresponding to the (two) solutions of said equation. Certain higher degree polynomial equations may still be solved using the quadratic formula, viz. (7.110) with n denoting an integer – after setting (7.111) that converts Eq. (7.110) to (7.112) Eq. (7.112) can be solved as discussed above to yield y 1 and y 2 as solutions. Reversion to the original nomenclature using Eq. (7.111) finally gives (7.113) and (7.114) In general, any equation of the form (7.115) may be solved for y 1 and y 2 via auxiliary variable (7.116) however, feasibility of analytically reversing to the original variable will depend on existence of an explicit form for f −1, according to (7.117) and (7.118) Although a solving formula exists for the cubic equation, it is far from being practical due to its outstanding complexity; hence, it is normally better to resort to a numerical method under such circumstances...
