Answer:
r = -12cos(θ)
Step-by-step explanation:
The usual translation can be used:
Putting these relationships into the formula, we have ...
(r·cos(θ) +6)² +(r·sin(θ))² = 36
r²·cos(θ)² +12r·cos(θ) +36 +r²·sin(θ)² = 36
r² +12r·cos(θ) = 0 . . . . subtract 36, use the trig identity cos²+sin²=1
r(r +12cos(θ)) = 0
This has two solutions for r:
r = 0 . . . . . . . . a point at the origin
r = -12cos(θ) . . . the circle of interest
Answer:
it will be very simple,when you understand the theory clerely, good luck!
1. When "a" is a root of a polynomial, (x -a) is a factor of it. For the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).
In order to make the leading coefficient be negative, we need to multiply this product by a negative number. Any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.
Then ...
... f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)
... = -(2x² +3x -2)(x -3)
... = -(2x³ -3x² -11x +6)
... f(x) = -2x³ +3x² +11x -6
2. A graph created by the Desmos on-line graphing calculator is shown, and the zeros are highlighted.
3. As indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. You want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.
