Find polynomial function p(x) such that p has the degree and p(x) = 0 has the roots listed

degree of p= 2
zeros= 1 + 6i

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Ответ:

emilypzamora11

26.01.2020

Mathematics

y = -0.75x + 9.5

Step-by-step explanation:

Equation of a circle is given as :

$(x-a)^{2} + (y-b)^2 = r^{2} \\(x-6)^{2} + (y-5)^2 = 4^{2}\\x^{2} - 12x + 36 + y^{2} -10y + 25 = 16\\x^{2} - 12x + y^{2} -10y + 45 = 0\\$

Take the derivative of the above with respect to x

$2x - 12 + 2y dy/dx - 10 dy/dy = 0\\\frac{dx}{dy} = \frac{6 - x}{y-5} \\At (2,8) dy/dx = (6-2)/(8-5) = 4/3\\dy/dx = 4/3\\The slope of the tangent to the circle, m = \frac{-1}{dy/dx} \\m = \frac{-1}{4/3}\\m= -0.75$

The equation of the tangent is given by :

$y - y_{1} = m(x - x_{1})\\ y - 8 = -0.75(x - 2)\\y - 8 = -0.75x + 1.5\\y = -0.75x + 9.5$

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Find polynomial function p(x) such that p has the degree and p(x) = 0 has the roots listed degree of p= 2 zeros= 1 + 6i