Q: In the standard equation of a circle centered at any point a change in value of the number that is part of the x-term results in a vertical movement?

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x2/4^2+y2/12^2=1

finding vertical asymptotes is easy. lets use the equation y = (2x-2)/((x^2)-2x-3) since its a rational equation, all we have to do to find the vertical asymptotes is find the values at which the denominator would be equal to 0. since this makes it an undefined equation, that is where the asymptotes are. for this equation, -1 and 3 are the answers for the vertical ayspmtotes. the horizontal asymptotes are a lot more tricky. to solve them, simplify the equation if it is in factored form, then divide all terms both in the numerator and denominator with the term with the highest degree. so the horizontal asymptote of this equation is 0.

vertical translations

A vertical line on a graph has an infinite slope, and no y-intercept.

No. The equation x=7 has an undefined slope since it is simply a vertical line located at x = 7. A basic test for a function is if you can draw a vertical line through the graph of the equation and it touches in more than one place, it is NOT a function.

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x2/4^2+y2/12^2=1

Ellipse formula, centered at the origin, where the vertical axis is the major axis: x2/b2 + y2/a2 = 1, a > b Since the major axis is 8, then a = 4. Since the minor axis is 4, then b = 2. Thus, the equation of the ellipse is: x2/4 + y2/16 = 1.