2 root 3x
y = -6x2 - 12x - 1We recognize this as the equation of a parabola opening downward, but we don't need to know that in order to answer the question.At the extremes of a function (local max or min), the first derivative of the function = zero.The first derivative of the given function with respect to 'x' is dy/dx = -12x -12Set -12x - 12 = 0.-x - 1 = 0x = -1y = -6x2 - 12x - 1 = -6(1) - 12(-1) - 1 = -6 + 12 - 1 = 5
The inverse operation to the derivative. Also called the integral. If you're given the derivative of a function, you can find the function again by performing the antiderivative. Many answers will be possible, all differing by a single number, so you normally add a general constant to the end. Example : The derivative of 6x^2 is 12x. The antiderivative of 12x is 6x^2 + any number.
y = x2 + 12x + 21At the max or min point, the first derivative of the function = 0.2x + 12 = 02x = -12x = -6
x^2 + 12x = 5 x^2 + 12x + __ = 5 + __ x^2 + 12x + 36 = 5 + 36 (The 36 I got from taking half of 12 and then squaring that number (which is 6).) (x + 6)^2 = 41 x + 6 = (the square root of 41) x = (the square root of 41) - 6. Hope that helped :)
Do you mean y2=x2-12x+17 Well you take the square root from both sides to get y by itself.
There is no answer to this problem unless x is 0. For the suare root of 98x to be a real number, x has to be positive or zero. For the square root of -147x to be a real number, x has to be negative or zero. Seeing has x has to fit both requirements, the problem has an answer only if x is zero.
Every polynomial defines a function, often called P. Linear and and quadratic function belong to a family of functions known as polynomial functions, which often are called P(x). When P(x) = 0, we call it an equation. Any value of x for which P(x) = 0 is a root of the equation and a zero of the function. Polynomials of the first few degrees have a special names such as:Degree 0: Constant functionDegree 1: Linear functionDegree 2: Quadratic functionDegree 3: Cubic functionDegree 4: Quartic functionDegree 5: Quintic functionSo, if we work a little bit to the given expression, we can turn it in a polynomial function of the second degree.y - 3x^2 = 12x - 7y - 3x^2 + 3x^2 = 12x - 7 + 3x^2y = 3x^2 + 12x - 7Let's write y = f(x) and f(x) = 3x^2 + 12x - 7, where a = 3, b = 12, and c = -7.Since a > 0, the parabola opens upward, so we have a minimum value of the function. The maximum or minimum value of the quadratic function occurs at x = -(b/2a).x = -12/6 = -2To find the minimum value of the function, which is also the y-value, we will find f(-2).f(-2) = 3(-2)^2 + 12(-2) - 7f(-2) = 12 - 24 - 7 = -19Thus the minimum value of the function is -19, and the vertex is (-2, -19)To find zeros, we solve f(x) = 0. So,f(x) = 3x^2 + 12x - 7f(x) = 03x^2 + 12x - 7 = 0 In order to solve this equation by completing the square, we need the constant term on the right hand side;3x^2 + 12x = 7 Add the square of one half of the coefficient of x to both sides, (6^2)3x^2 +12x + 36 = 7 + 36 Use the formula (a + b)^2 = a^2 + 2ab + b^2;(3x + 6)^2 = 43 Take the square root of both sides, and solve for x;3x + 6 = (+ & -)square root of 433x + 6 = (+ & -)(square root of 43) Subtract 6 to both sides;3x = (+ & -)(square root of 43) - 6 Divide both sides by 3;x = (square root of 43)/3 - 2 or x = -(square root of 43)/3 - 2The solution are (square root of 43) - 2 and -(square root of 43) - 2
√(12x3y3z-2) * √(15xy) = √(180x4y4x-2) = 13.416x2y2z-1 = 13.416(xy)2/z
d/dx(X^4) = 4X^3 ( first derivative ) d/dx(4X^3) = 12X^2 ( second derivative )
-0.2x2+12x+11 = 0 Using the quadratic equation formula will give you two solutions which are: x = 30 + the square root of 955 = 60.90307428 to 8 decimal places. or: x = 30 - the square root of 955 = -0.90307428 to 8 decimal places.