Write a function whose graph is a hyperbola that has the given asymptote and passes through the point.

Question

1.) x=5, y=6;(4,2)

 

2.) x= -1 y=-3; (1,2)

 

 

Answer 1

The Graphing Form for the equation of a hyperbola function is:

 

y = a (1 / (x-h)) + k ;  (x,Y) is a point on the graph, (h,k) is called

                                   the locator point

 

The locator point could be described as the point of intersection of the

asymptotes

 

Asymptotes: x = 5 , y = 6  ; point (4,2) is on the curve

 

y = a (1/(x - 5)) + 6

 

insert (4,2) into equation and solve for a

 

      2 = a (1/ (4 - 5)) + 6

 

     -4 = a(1/-1) 

 

      4 = a

 

Equation is:   y = 4/ (x - 5) + 6

 

The other problem follows this same pattern

Answer 2

The hyperbolas with transverse axis parallel to x-axis or to y-axis have asymptote equations of the form

y-y₁ = ±m(x-x₁)  where m is either b/a or a/b and (x₁,y₁) is the center, etc.

 

The x=5 & y=6 cannot be equations of asymptotes, in this case. (They could only be asymptotes if the hyperbola were rotated, and that's not an Algebra 1 or Rational Functions type of problem.)

 

So I think that something is wrong in the statement of the problem.