Write an equation of the ellipse with foci at (0, +2) and co-vertices at (+1,0

Step-by-step explanation:

We are given the foci of the ellipse as (0, 2) and (0, -2).  "c" is the distance from the center to one focal point.  Our center is directly between the 2 foci, so the center is located at (0, 0).  "c" then is 2.

We are also given the co-vertex of (1, 0) and (-1, 0) which is located on the minor axis (shorter axis).  The foci ALWAYS LIE ON THE MAJOR AXIS, and if we just determined that the shorter axis is the x axis (because the points (1, 0) and (-1, 0) lie on the x-axis) then our ellipse is vertically stretched.  The equation for a vertically stretched ellipse is

As far as the "a" and the "b" go, a is ALWAYS LARGER THAN B, AND WILL ALWAYS LIE UNDER THE AXIS THAT IS THE MAJOR AXIS.  Y is our major axis, therefore, a lies under the y-squared fraction.  (Also, in an ellipse, the x-squared and the y-squared do not move...only the a and the b do.  In a hyperbola, the a and b remain fixed and the x-squared and y-squared move.)

So we know that c = 2 and b = 1, now we need to find a.  The equation for the foci of an ellipse is

Solving this for a:

so

and

Filling in our equation now with a and b:

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

the cat is blue because the probability of extension towards the prefix of the diameter of the circle could only mathematically equal to the sum of the roots of the carrots from   your garden.

step-by-step explanation: