Wikipedia
Quadratically constrained quadratic program - In mathematics, a quadratically constrained quadratic program ( QCQP) is an optimization problem in which both the objective function and the constraints are quadratic functions. It has the form& ext{minimize} && frac12 x^ op P_0 x + q_0^ op x \ & ext{subject to} && frac12 x^ op P_i..
Quadratically constrained quadratic program - In mathematics, a quadratically constrained quadratic program ( QCQP) is an optimization problem in which both the objective function and the constraints are quadratic functions. It has the form #8;egin{align} & ext{minimize} && frac12 x^ op P_0 x + q_0^ op x \ & ext{subject..
TutorVista
Objective Function
The Objective Function is a linear function of variables which is to be optimised i.e., maximised or minimised. e.g., profit function, cost function etc. The objective function may be expressed as a linear expressio..
Limits on the variables in the objective function are called _______.
Limits on the variables in the objective function are called _______. => Extremes or Vertices or Restrictions or Objectives..
Determine which of the graphs represent a quadratic function.
Determine which of the graphs represent a quadratic function. => Graph 1 and Graph 4 or Graph 1 only or Graph 1 and Graph 3 or Graph 4 only..
Find the quadratic function which has the zeros 4, - 9.
Find the quadratic function which has the zeros 4, - 9. => f ( x ) = x 2 - 5 x - 36 or f ( x ) = x 2 + 5 x - 36 or f ( x ) = x 2 + 5 x + 36 or f ( x ) = x 2 - 5 x + 36..
How do you graph a quadratic function?how to graph a quadratic function
Quadratic Functional GraphQuadratic Functional Graph - parabola
Question : For each quadratic function given below, find its vertex, axis of symmetry, y-intercept and x-intercept, if any. Determine the domain and range of each function. Determine where the functions are increasing and where they are decreasing.
(1) f(x) = x^2 - 4x
(2) f(x) = -3x^2 + 3x - 2
Please, do not answer the questions unless you know how to easily explain what to do. Do not use math jargon. Clear?
Answer : Problem #1 ---------------------- f(x) = x^2 - 4x a) To find the vertex, put the equation into the form: y=(x-x0)^2 + y0 This can be done by 'completing the square': f(x) = x^2 - 4x + 4 - 4 f(x) = (x-2)^2 - 4 From this form, we can directly see that the vertex is at: (2,-4) b) From the form of the equation, it is obviously a parabola (upward). Since the vertex is at (2,-4), we can see that the axis of symmetry is 'x=2' c) To find the y-intercept, set x to 0: y = x^2 - 4x y = 0 - 0 y = 0 d) To find the x-intercept, set y to 0: y = x^2 - 4x 0 = x^2 - 4x 0 = x(x-4) Therefore, x-intercept at x=0 or x=4 e) The domain (legal x values) is all Reals. The range (possible y values) is y>=-4. f) Since it is a upward facing parabola, we can easily see: decreasing for -inf < x < 2 increasing for 2 < x < inf More formally, however, if we take th..
Answer : Problem #1 ---------------------- f(x) = x^2 - 4x a) To find the vertex, put the equation into the form: y=(x-x0)^2 + y0 This can be done by 'completing the square': f(x) = x^2 - 4x + 4 - 4 f(x) = (x-2)^2 - 4 From this form, we can directly see that the vertex is at: (2,-4) b) From the form of the equation, it is obviously a parabola (upward). Since the vertex is at (2,-4), we can see that the axis of symmetry is 'x=2' c) To find the y-intercept, set x to 0: y = x^2 - 4x y = 0 - 0 y = 0 d) To find the x-intercept, set y to 0: y = x^2 - 4x 0 = x^2 - 4x 0 = x(x-4) Therefore, x-intercept at x=0 or x=4 e) The domain (legal x values) is all Reals. The range (possible y values) is y>=-4. f) Since it is a upward facing parabola, we can easily see: decreasing for -inf < x < 2 increasing for 2 < x < inf More formally, however, if we take th..
Question : Prove that the roots of the equation:
x^2 + (k + 2)x + 2k = 0
are real for all values of k.
I need help on this, the last set of answers where wrong,
all i know that-
Real roots: b^2 > 4ac
Anyone online that can help?
Answer : So just write down what you already wrote (only it is slightly wrong: it needs a >= instead of just a >): The equation has real roots whenever b >= 4ac Now plug in and simplify: The equation has real roots whenever (k+2) >= 4(1)(2k) k +4k+4 >= 8k k -4k+4 >= 0 (k-2) >= 0 Since the term on the left is a square, it is always greater than or equal to 0. Thus, you will always have real roots.
Answer : So just write down what you already wrote (only it is slightly wrong: it needs a >= instead of just a >): The equation has real roots whenever b >= 4ac Now plug in and simplify: The equation has real roots whenever (k+2) >= 4(1)(2k) k +4k+4 >= 8k k -4k+4 >= 0 (k-2) >= 0 Since the term on the left is a square, it is always greater than or equal to 0. Thus, you will always have real roots.