Transformations

For non-linear, HL transformations, see composition of function and absolute value, and reciprocal and square of functions.

While functions are like families of equations, transformations of functions are like families of functions.

Transformations allow us to examine related functions without doing too much work. In particular, we shall apply linear transformations to xx and yy, such that functions are stretched, reflected, or translated along the horizontal and vertical directions.

Contents

Everything will make sense (I promise)

This derivation is very good to know, but it is not asked on exams.

Say we are familiar with some function y1=f1(x1)y_1 = f_1(x_1), and we want to learn more about some related y2=f2(x2)y_2 = f_2(x_2). The functions are related such that

x2=bx1+hx_2 = bx_1 + h
y2=ay1+ky_2 = ay_1 + k

Our goal is to rewrite f2f_2 using f1f_1, and we want to evaluate f1f_1 using some expression of x1x_1. An analogy is that if we are switching the locks, then we also have to switch the keys.

x1=1b(x2−h)x_1 = \frac 1b (x_2 - h)

Then using y1=f1(x1)y_1 = f_1(x_1), we obtain

y2=af1(1b(x2−h))+ky_2 = af_1\left(\frac 1b (x_2 - h)\right) + k

We shall refer ∣b∣\lvert b\rvert and ∣a∣\lvert a\rvert as the horizontal and vertical stretch factors. If bb and aa are negative, then there is also a reflection across the yy- or xx-axis. Finally hh and kk are horizontal and vertical translations, which are applied after stretches and/or reflections.

Note: a horizontal reflection is across the vertical line x=0x = 0 (yy-axis), and vice versa.

In other words, when expressing f2f_2 as a transformation of f1f_1, we apply the xx-transformations in reverse, evaluate through f1f_1, then apply the yy-transformations in the forward order.

HL candidates should be able to rewrite f(3x+5)f(3x + 5) as f(3(x+53))f\left(3\left(x + \frac{5}{3}\right)\right) to identify the horizontal stretch factor 13\frac13 and horizontal shift of 53\frac53 units to the left.

Theorem: an xx- (or yy-) transformation corresponds to a change in the domain (or range).

Note: a compression factor is just when the stretch factor is between 00 and 11. Eg if the stretch factor is 12\frac12, it can also be said that the compression factor is 12\frac12, to emphasize that the graph is squeezed together. There is no sign change or reciprocal going on.

Examples

Example: f(x)=16−x2f(x) = \sqrt{16 - x^2} is a semicircle with radius 44 and center at (0,0)(0, 0). Find the domain and range of g(x)=−316−4x2−5g(x) = -3\sqrt{16 - 4x^2} - 5.

yy-transformations are easier to see. We see a vertical stretch factor of 33, a vertical reflection across xx-axis, and a translation 55 units down.

The range of ff is [0,4][0, 4]. The range of g(x)g(x) is

[−3(0)−5,−3(4)−5]=[−5,−17]=[−17,−5]■\left[-3(0) - 5, -3(4) - 5\right] = \left[-5, -17\right] = \left[-17, -5\right] \qed

For xx-transformations, we have to re-write gg to involve transformation, with g(x)=−316−(2x)2−5{g(x) = -3\sqrt{16 - (2x)^2} - 5}

This corresponds to a horizontal stretch by a factor of 12\frac{1}{2}.

The domain of ff is [−4,4]\left[-4, 4\right]. The domain of g(x)g(x) is [−2,2]■\left[-2, 2\right] \qed

Example: The graph y=f(x)y = f(x) contains the point (1,−2)(1, -2).

A second function gg is f(x)f(x) vertically stretched by a factor of 4, horizontally reflected across the yy-axis, horizontal stretched by a factor of 33. then translated 5 units up and 6 units to the left.

a) Find the corresponding point on y=g(x)y = g(x).

b) Find an expression for gg in terms of ff.

a)

(−3(1)−6,4(−2)+5)=(−9,−3)■(-3(1) - 6, 4(-2) + 5) = (-9, -3) \qed

b)

g(x)=4f(−13(x+6))+5■g(x) = 4f\left(-\frac13(x+6)\right) + 5 \qed

Note: Applying xx-transformation on the xx-coordinate applies the transformation in the forward direction, whereas the (1b(x−h))\displaystyle \left(\frac1b(x - h)\right) applies it in reverse.