2.05 Transformations of hyperbolas

Transformations - dilation

Use the following applet to explore the effect that $$a has on the hyperbola $$y=ax​. Adjust the values of $$a and try to summarise the effect.

Summary:

• $$a dilates (stretches) the graph by a factor of $$a from the $$x-axis
• The larger the magnitude of $$a the further the graph is from the origin
• The point $$(1,1) will be stretched to the point $$(1,a)
• When $$a is negative the graph lies in the $$2nd and $$4th quadrants. This is a reflection of the graph $$y=|a|x​ in the $$x-axis.
• The graphs still has the same symmetry properties of $$y=1x​
• The graph has a vertical asymptote of $$x=0 and a horizontal asymptote of $$y=0

Can you find the coordinates of the 'corner' point which is the closest point to the origin? Hint: It lies on the line $$y=x.

 

Transformations - translation

Use the next applet to explore the effect that $$h and $$k have on the hyperbola $$y=ax−h​+k. Adjust the values of $$h and $$k and try to summarise the effect.

Summary:

• The hyperbola given by $$y=ax−h​+k can be thought of as the basic rectangular hyperbola $$y=ax​ translated horizontally (parallel to the $$x axis) a distance of $$h units and translated vertically (parallel to the $$y axis) a distance of $$k units
• The centre (intersection of the asymptotes) will move to the point $$(h,k)
• The orientation of the hyperbola will remain unaltered
• Vertical asymptote is translated to $$x=h
• Horizontal asymptote in translated to $$y=k

For example, to sketch the hyperbola $$y=12x−3​+7, first place the centre at $$(3,7). Then draw in the two orthogonal asymptotes (orthogonal means at right angles) given by $$x=3 and $$y=7. Finally, draw the hyperbola as if it were the basic hyperbola $$y=12x​ but now centred at the point $$(3,7).

Note that the domain includes all values of $$x not equal to $$3 and the range includes all values of $$y not equal to $$7.

 

Domain and range

We have seen that the function $$y=ax−h​+k has asymptotes given by $$x=h and $$y=k. Thus $$x=h is the only point excluded from the domain and $$y=k is the only point excluded from the range.

We usually state this formally as, in the case of the domain, $$x:x∈ℝ,x≠h and in the case of the range, $$y:y∈ℝ,y≠k. Alternatively, we can use interval notation, then the domain can be written as $$(−∞,h)∪(h,∞). And the range can be written as $$(−∞,k)∪(k,∞).

Rather than thinking of translations we can also see from the equation that the domain and range exclude these values. From the form $$y=ax−h​+k, we can see that $$x=h would cause the denominator to be zero and hence, the expression to be undefined. We can rearrange the equation to either $$y=ax−k​+h or $$(x−h)(y−k)=a, to see that $$y=k will also cause the equation to be undefined.

 

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3