State the type of the quadratic surface: x2+(y6)2+z2=1 1. hyperboloid of two sheets 2. hyperboloid of one sheet 3. ellipsoid 4. none of these

solution:

quadratic surface,

x^{2}+z^{2}+\frac{y^{2}}{36}=1

=x^{2}+z^{2}=1-\frac{y^{2}}{36}

=\frac{x^{2}}{a^{2}}+\frac{z^{2}}{c^{2}}=1-\frac{y^{2}}{b^{2}}

this is a hyperbolid of one that intersection at  

y=0

the quadratic surface becomes

x^{2}+z^{2}=1

this is the equation of the unit circle

so,the surface is three circle

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: