Parallel displacement and the functions of parabolic and hyperbolic growth

The general equation of the quadratic function is y = x² + ax + b. As you inspect this graph for a = 4 and b = 3, you note that the curve is like that of y = x²; in fact, the one curve arises from the other by parallel displacement. This suggests that you can by such displacements find all curves y=x²+ax+b. Hence we will apply several such displacements to it and find out what they do to the graph of y = x².

Displacement along the ordinate axis by b units

In the next figure with b=1, every point of y=x² has been moved vertically upwards by one unit. The abscissae remain the same, the new ordinates y' are by 1 larger than y. Hence, in order to obtain y=x² from y', you must subtract 1 from all its values. Then its new equation is y' - 1 = x'². It appears in an implicit form (cf. Concept of function), but that does not matter. In order to make it explicit, you add +1 to both sides of the equation and arrive at y' =x'² + 1. In the next figure b = - 2, so that it is y'+2=x'² and you must add 2 to obtain the ordinates of y = x².

Displacement along the abscissa by a units

During displacement of y=x² along the abscissa, the ordinates remain the same and the abscissae change; in the next figure above, they change by a = +2. Hence you must always subtract 2 from the abscissa x', in order to obtain the old values of x. The equation of the graph is y' = (x' - 2)². In the next figure above, it is y' = (x' +1)².

Displacement along both axes

Displacement of the parabola by a=+2 leads to the equation y=(x - 2)²; if we displace this curve by b = -3 along the ordinate axis, we obtain the equation y + 3 = (x - 2)².

These manipulations of the function y=x² can be applied to any function. If you introduce for all functions, for example, for y=x² or y=x³, etc. the notation y = f(x), you have the Rule:

The graph of the function y = f(x) after displacement along the abscissa by a units and along the ordinate by b units has the equation y - b = f(x - a).

In order to draw other graphs of y = x², it is useful to make it into a pattern. It is also useful to conceive displacements of the coordinate system. If you shift the origin O by a units along the x-axis and b units along the y-axis, the image O' of the origin O must have the coordinates (a,b) and the new axes O'x' and O'y' must be parallel to the axes Ox and Oy. You first mark the origin, place axes parallel to the Oxy-axes and then place your pattern, corresponding to the normal position. The equations of the curve remain, of course, always referred to the original coordinate system.

It is now important to deal with the reverse task. Given an equation, graph the curve. If you can give it the form y - b = f(x - a), then you can solve the problem by parallel displacement of the curve.

y = ax² + bx + c, the function of parabolic growth.

In the beginning, let us start with the normal form y = ax² + bx + c and solve the Task:

Draw the curve y=x²+6x+11.

You will probably try to reduce the given equation to the form y - b = (x - a)², used above. Add to both sides of the equation y - 11= x² + 6x a number, so that there will be a square expression on the right hand side. Comparing x²+6x with a² + 2ab + b², you conclude that, if a=x, 2bx=6x, then b = 3 and b² = 9. This so called quadratic complement is the square of half the coefficient of x. Addition of 9 to both sides of the equation y - 11= x² + 6x yields y - 2 = (x + 3)². The origin of the new coordinate system has the coordinates a = -3, b = +2. You use now your pattern of y=x² at the right location and obtain the curve sought.

By parallel displacement of the parabola y=x², you can graph every equation y=ax²+bx+c; the proposal made at the start of this section has been confirmed.

If you now start from the curve y = Ax² (A 0), a result of stretching of y=x² along the ordinate axis (cf. Graph of the function y = axⁿ), and apply to it a parallel displacement, the resulting curve has the equation y - b = A(x - a)². Into this form you can again transform every equation of the general form y = Ax² + Bx + C.

Task: Draw the curve y = 2x² + 16x + 33!

You have y -33= 2(x² + 8x). The quadratic complement of
x² + 8x is the square of half the coefficient of x, that is 4²=16. If you add inside the brackets 16, then, since there is a 2 in front of it, you must add 32 to the left hand side of the equation:
y - 33 + 32 = 2(x² +8x + 16 or y-1=2(x+4)².

In practice, you first displace the parabola y=x² and then stretch it. Since you have here the coefficient 2, you double the ordinates of the displaced parabola with respect to the abscissa; you can also proceed as has been shown in Graph of the function y = axⁿ.

As one speaks in connection with the function y = mx + n of linear growth, you can in connection with an equation of the form y = Ax² + Bx + C speak of parabolic growth.

y = Ax² + Bx + C is the function of parabolic growth

In a similar manner as above, you can arrive at the graphs of other functions.

Task: Obtain the graph of the function .

The curve becomes through parallel displacement , whence we find Seek the point O' (a = 0, b = +4) and make now a pattern of the parabola in the form and then stretch it, using the earlier method (cf. Graph of the function y = axⁿ). You can obtain from a right-angled triangle with two equal sides of length 1 cm!

y = (ax + b)/(cx + d), the function of hyperbolic growth

Task: Graph the function y = (2x + 5)/(x + 2).

Divide out on the right hand side: (2x + 5) : (x + 2) = 2 + 1/(x + 2), whence the given function becomes y = 2 + 1/(x + 2) or y - 2 = 1/(x + 2). It has now the form y - b = m/(x - a), which one obtains by parallel displacement of the hyperbola y = m/x. In this case, m = 1, a = -2, b = +2. You have to move the hyperbola y = 1/x along the abscissa by -2 units and along the ordinate by + 2 units (cf. the fourth last figure). Since you can reduce every such function in the same manner to a hyperbola, we speak here of the function of hyperbolic growth.

It is possible that the division does not leave a remainder, that is, that the numerator is a multiple of the denominator cx + d. Then the equation becomes y = n(cx + d)/(cx + d), whence ax + b = n(cx + d) or y(cx + d) = n(cx + d) or ycx + yd = ncx + nd or cx(y - n) + d(y - n) = 0 or (y - n)(cx + d) = 0. Then either y - n = 0 for any x or cx + d =0 for any y. The hyperbola degenerates into 2 straight lines (its asymptotes). c cannot be zero, because the function is not linear.

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