Ahyperbola is centered at the origin and opens either horizontally or vertically. it passes through the points (-3, 4), (-2, 0), and (t, 2). find t^2.

  5.25

Step-by-step explanation:

The fact that the graph crosses the x-axis means the hyperbola opens to the left and right. Hence its equation is ...

  x^2/a - y^2/b = 1

The point (-2, 0) helps us find "a":

  (-2)^2/a - 0/b = 1

  4 = a

Then we can find "b" from the other fully-specified point.

  (-3)^2/4 - 4^2/b = 1 . . . . . . substitute (-3, 4) for (x, y)

  b = 16/(9/4 -1) = 64/5 . . . . solve for b

Then the third point will satisfy ...

  t^2/4 - (5/64)·2^2 = 1 . . . . . . substitute (t, 2) for (x, y)

  t^2 = 4·(1 +5/16) = 4 +5/4 . . . . solve for t^2

  t^2 = 5.25