Poles at infinity are obtained when the order of the numerator is higher than the order of the denominator. Consider a transfer function G(s) with a numerator of order n, and denominator of order m, and with n>m. There will be n finite zeros and m finite poles and, as s->infinity, the m poles will cancel m of the numerator zeros leaving (n-m) zeros, therefore G(s) -> infinity and there will be (n-m) poles at infinity.
e.g. G(s)=(s+a) has a finite zero at s=-a and a pole at s=infinity
Conversely, if the denominator is of higher order than the numerator (as must be the case for a physically realisable system), m>n, then there will be (m-n) zeros at infinity.
e.g. G(s)=1/(s+b) has a finite pole at s=-b and a zero at s=infinity; G(s)=(s+a)/s(s+b) has finite poles at s=0 and s=-b, a finite zero at s=-a, and a zero at s=infinity
In terms of design, the root locus often finds application. This typically tracks the locus of the closed-loop poles as the forward-path gain, K, is increased from zero to infinity. The locus may have several branches and these start at the finite open-loop poles and end at the open loop zeros; but if there are more poles than zeros (as is usual), the excess branches that cannot find a finite zero to terminate at will move to zeros at infinity.
It's enlightening to determine the frequency response graphically using the complex s-plane. To do this the finite poles and zeros are plotted, then to find the gain (and phase angle) at any given frequency, w, draw vectors from the point s=jw on the imaginary axis, to all the poles and zeros and the gain will be the product of the lengths of the zero vectors divided by the product of the lengths of the pole vectors. The phase angle will be the sum of all the angles made by the zero vectors minus the sum of all the angles made by the pole vectors. If there are no finite zeros (e.g. G(s)=K/(s+a)), then just use K for the vector length and 0 for the angle.
So, for example, if there is a pole on the jw axis at, say, s=jw1, then as w increases from 0, the length of the vector from s=jw to s=jw1 will decrease until, when w = w1, the length of the vector will be zero and the gain of the TF will be infinite. This is resonance and, since the pole is on the imaginary axis, the response is infinite at this frequency. If the pole is slightly off the imaginary axis the vector length will not reach zero, but will bottom-out as the trajectory passes it on its journey up the jw axis. This too is resonance, but the resonance peak is not of infinite height as the vector length never reaches zero. The further the pole is from the jw axis, the lower is the resonance peak. i.e. as poles move to the left, away from the jw axis the system resonances become lower in magnitude - the system becomes more stable (and faster - but that's a another story)