Standard form of a circle" (x-h)²+(y-k)²=r², (h,k) being the center, r being the radius.
in this case, h=-2, k=6, (x+2)²+(y-6)²=r²
use the point (-2,10) to find r: (-2+2)²+(10-6)²=r², r=4
so the equation of the circle is: (x+2)²+(y-6)²=4²
Answer:
F(-1) = 1
x = 3
Step-by-step explanation:
1) Assume that f(-1) = y1
So that point A (x1 = -1; y1) is a point lying in the graph representing the equation y = f(x)
As it can be seen in the figure, the graph representing that equation crosses point (-1; 1)
=> Point A has y1 = 1
=> f (-1) =1
2) Assume that f(x2) = -2
So that point B (x2; y2 = -2) is a point on the graphy representing the equation y = f(x)
As indicated in the figure, the graph crosses point (3; -2)
=> Point B has x2 = 3
=> When f(x) = -2, x = 3
We are asked in the problem to devise a polynomial equation that has a GCF of 6 which means each of the terms can be divided to 6. For example: 6*(x^2 + x+1) = 6x^2 + 6x +6. This polynomial is created by multiplying each terms by the number 6 which is distinguished by factoring.
