Mathematics
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.
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eBook - ePubDifferentiating Instruction in Algebra 1
Ready-to-Use Activities for All Students (Grades 7-10)
- Kelli Jurek(Author)
- 2021(Publication Date)
- Routledge(Publisher)
Directions: Choose three problems in a row, in a column, or on a diagonal. 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 + bx2 + 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 - r1 )(x - r2 ), where r1 and r2 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.
When you have completed the problems, check your answers. Each box is worth 1 point.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 = 2x2 + 4x - 6 y = x2 - 3x + 7 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
- Lawrence S. Leff, Christina Pawlowski-Polanish(Authors)
- 2021(Publication Date)
- Barrons Educational Services(Publisher)
c = 25, soSimilarly, if ax2 + 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 ax2 + bx + c = 0, in which the roots of the equation are expressed in terms of the coefficients a, b, and c.QUADRATIC FORMULAIf ax2 + bx + c = 0 and a ≠ 0, thenAny quadratic equation, including equations with rational, irrational, or imaginary roots, can be solved by writing the equation in the form ax2 + 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 FormulaFind the roots of 2x2 − 3x − 1 = 0 in radical form.SOLUTIONUse the quadratic formula, where a = 2, b = −3, and c = −1:If the two roots of a quadratic equation are denoted by x1 and x2 , thenAlthough 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 EquationFind the roots of x2 + 5 = 2x in a + bi form.SOLUTIONIf x2 + 5 = 2x, then x2 − 2x + 5 = 0. Use the quadratic formula, where a = 1, b = −2, and c = 5:Hence, x1 = 1 + 2i and x2 = 1 – 2i.MAKING GRAPHICAL CONNECTIONS
The expression b2 − 4ac underneath the radical sign in the quadratic formula is called the discriminant. The discriminant determines the nature of the roots of a quadratic function as well as the number of x-intercepts of its related graph. See the accompanying table where a, b, and c
eBook - ePub- Stu Schwartz(Author)
- 2013(Publication Date)
- Research & Education Association(Publisher)
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 2b. f (x ) = x 2 − 4c. f (x ) = x 2 + 1SOLUTION
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) = 0x + 2 = 0 x − 2 = 0x = −2 or x = 2Note 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. If we set the function equal to zero, we get x 2 = −1, which has no real solutions. We will discuss this further in Chapter 6
eBook - ePub- Michael Harrison, Patrick Waldron(Authors)
- 2011(Publication Date)
- Routledge(Publisher)
Conic sections, quadratic forms and definite matrices
DOI: 10.4324/9780203829998-6 f4.1 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.1Consider the curve traced out in the coordinate plane ℝ2 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. The equation of the parabola takes its simplest form when the focus S is a point on the positive x axis, say (a, 0), where a > 0, and the directrix is the vertical line with equation x = −a. An example of a parabola is shown in Figure 4.1
- 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
