1. From the equation, there are asymptotes at x = 3 and y = 0. Draw these. There is no sign to reverse the x term. So a normal hyperbola.

2. From the equation, there are asymptotes at x = -2 and y = 0.
Draw these. There is a negative sign in from of the x fraction so the hyperbola is flipped.

4. Add the "3" to both sides gives
y + 3 so there is a horizontal asymptote when y + 3 = 0 or
at y = -3. Similarly there is a vertical asymptote when 2 - x = 0
so at x = 2.

Although there appears to be a + sign in front of the fraction, it is actually a negative when the terms are reversed. (i.e. -(x - 2)). So the hyperbola is flipped vertically.

Determine the equation from a graph.

5.

The curve is balanced on either side of the y axis and so there is no horizontal transformtion of the basic parabola.

There is an asymptote at y = 1 so the hyperbola is shifted up 1.

Hence the equation is:

The shape of this graph is very similar to the shape of that in Q5. The asymptote at y = 1 is still there (so the equation must start
" y = 1 + ". The hyperbola has now also been shifted to the left by 1 - so we use x + 1.

Hence the equation is

This hyperbola has been shifted from the basic position 3 units to the right and 2 units down. So write these shifts in terms of where the asymptotes are now:

x = 3 and y = -2

Now rewrite as x - 3 = 0 and y + 2 = 0
and substitue into the basic hyperbola equation:

The basic hyperbola has been flipped vertically about the y axis and then shifted from its basic position 2 units to the right and 1 units up. So write these shifts in terms of where the asymptotes are now: x = 2 and y = 1.

Then rewrite as x - 2 = 0 and y - 1 = 0 as usual.
Substitute into the basic hyperbola equation and then change the sign in front of the fraction containing the x term from +ve to -ve. Then tidy up.

9. For the hyperbolic equation ,

there are asymptotes when x - 1 = 0 -
so at x = 1 and at y = 0.

The absolute value leaves the RH arm of the hyperbola unaltered but the LH arm (which used negative values of y) is flipped around the x axis to become positive.

10. For the hyperbolic equation
, do nothing different to our standard procedure.

2x - 5 = 0 shows an asymptote at x = 2½ and y - 2 shows an asymptote at y = 2.

Now draw it.

(so the 2 in front of the x makes no difference to our technique).

If there was a number greater than 1 in the numerator, that will simply push the arms away from the intersection of the asymptotes. How far can be determined by solving the equations for the axis intercepts.

(b The equation can be analysed:

x + 1 = 0 gives a vertical asymptote
at x = -1.
y - 3 = 0 shows an horizontal asymptote
at y = + 3.

There is a positive sign in front of the fractional x term - so no flipping:

(b) Hence or otherwise, sketch a graph of Showing all important features.

(b) The x term has no number attached so the y axis is an asymptote.

subtracting gives y - 3 on the left side so there is a horizontal asymptote at y = 3.

The fractional x term is negative so the hyperbola is flipped vertically.