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How can you use a graph to find zeros of a quadratic function?
The zeros of a quadratic function, if they exist, are the values of the variable at which the graph crosses the horizontal axis.
How do you find all the zeros of a function?
You could try setting the function equal to zero, and finding all the solutions of the equation. Just a suggestion.
What is next number is series 298 209 129 58 - 4?
The answer can be any number that you like: it is always possible to find a polynomial of order 5 to fit the given numbers and any other number.The lowest degree polynomial that will fit the given numbers is the quadraticUn = (9n2 - 205n + 792)/2 for n = 1, 2, 3, .. . and that gives the next number as -57.The answer can be any number that you like: it is always possible to find a polynomial of order 5 to fit the given numbers and any other number.The lowest degree polynomial that will fit the given numbers is the quadraticUn = (9n2 - 205n + 792)/2 for n = 1, 2, 3, .. . and that gives the next number as -57.The answer can be any number that you like: it is always possible to find a polynomial of order 5 to fit the given numbers and any other number.The lowest degree polynomial that will fit the given numbers is the quadraticUn = (9n2 - 205n + 792)/2 for n = 1, 2, 3, .. . and that gives the next number as -57.The answer can be any number that you like: it is always possible to find a polynomial of order 5 to fit the given numbers and any other number.The lowest degree polynomial that will fit the given numbers is the quadraticUn = (9n2 - 205n + 792)/2 for n = 1, 2, 3, .. . and that gives the next number as -57.
What is linear in 4x2 3x - 6?
4x2 + 3x - 6 is a second degree polynomial. Since the polynomial function f(x) = 4x2 + 3x - 6 has 2 zeros, it has 2 linear factors. Since we cannot factor the given polynomial, let's find the two roots of the equation 4x2 + 3x - 6 = 0, which are the zeros of the function. 4x2 + 3x - 6 = 0 x2 + (3/4)x = 6/4 x2 + (3/4)x + (3/8)2 = 6/4 + 9/64 (x + 3/8)2 = 105/64 x + 3/8 = ± √(105/64) x = (-3 ± √105)/8 x = -(3 - √105)/8 or x = -(3 + √105)/8 Thus, the linear factorization of f(x) = 4[x + (3 - √105)/8][x + (3 + √105)/8].
How do you find a 4th degree polynomial with zeros x equals 4 x equals -3 and x equals 6i?
There are only three roots given so, in general, there is no unique answer. However, if it is a real polynomial, then its complex roots must come in conjugate pairs. Then 6i is a root implies that -6i is a root. So the polynomial is (x - 4)(x + 3)(x + 6i)(x - 6i) = (x2 - x - 12)(x2 + 36) = x4 + 36x2 - x3 - 36x - 12x2 - 432 = x4 - x3 + 24x2 - 36x - 432
