A. Exponential and Logarithmic Functions · Exponential and Logarithmic Functions

Transforming Logarithmic Graphs I: Translations, Stretches and Reflections

A2.3About 70 minutes

Learning goals

I can describe how c and d change the graph of y = log (x − d) + c.
I can describe how a changes the graph of y = a log x, including a reflection when a < 0.
I can sketch y = a log (x − d) + c by transforming the key points (0.1, −1), (1, 0) and (10, 1) and the asymptote.
I can state the domain, range and asymptote of a transformed logarithmic function.

Words to know

parent function
The basic function a family is built from. Here it is f(x) = log x, with key points (0.1, −1), (1, 0) and (10, 1).

horizontal translation
A shift left or right. In y = log (x − d), the graph moves d units right (left if d < 0), and so does the asymptote.

vertical translation
A shift up or down. In y = log x + c, the graph moves c units up (down if c < 0).

vertical stretch or compression
In y = a log x, every y-value is multiplied by a. |a| > 1 stretches, 0 < |a| < 1 compresses.

reflection in the x-axis
When a < 0, every y-value changes sign: the graph flips over the x-axis.

The same toolbox, a new function

Remember from Grade 11: you transformed parabolas, square roots, reciprocals and exponential functions with y = af(k(x − d)) + c. The same four parameters work for f(x) = log x. In this lesson you will use a, d and c; in lesson 9 you will add k.

The key points of the parent function y = log x are (0.1, −1), (1, 0) and (10, 1), and its asymptote is x = 0. Transform these and you can sketch any member of the family.

If you have graphing technology (for example the free Desmos calculator), type y = a log(x − d) + c, add sliders and watch what each parameter does. Then confirm by hand using the key points, as in the examples below.

Translations: c and d

y = log x + c moves the graph up c units (down if c < 0). The asymptote x = 0 does not move.
y = log (x − d) moves the graph right d units (left if d < 0). The asymptote moves too: it becomes x = d, and the domain becomes x > d.

Watch the sign: y = log (x − 3) moves right 3, while y = log (x + 3) = log (x − (−3)) moves left 3.

y = log (x − 3) is y = log x moved 3 right, so its asymptote is x = 3 and (1, 0) moves to (4, 0). y = log x + 2 is y = log x moved 2 up: (10, 1) moves to (10, 3).

Vertical stretches and reflections: a

In y = a log x, multiply every y-value by a. The x-values, the domain and the asymptote do not change. The x-intercept stays at 1, because a × 0 = 0.

|a| > 1: vertical stretch (steeper).
0 < |a| < 1: vertical compression (flatter).
a < 0: also a reflection in the x-axis (the graph decreases).

Multiplying by 2 doubles every y-value: (10, 1) becomes (10, 2). Multiplying by −1 reflects in the x-axis: (10, 1) becomes (10, −1). All three graphs keep the point (1, 0) and the asymptote x = 0.

Mapping rule

For y = a log (x − d) + c, each point (x, y) on y = log x moves to (x + d, ay + c). Do the stretch or reflection first, then the translations. The asymptote x = 0 becomes x = d.

Worked example

Example 1: Translations

Sketch y = log (x − 3) + 2. State its domain, range, asymptote and x-intercept.

Mapping rule: (x, y) → (x + 3, y + 2).
(0.1, −1) → (3.1, 1); (1, 0) → (4, 2); (10, 1) → (13, 3).
Asymptote: x = 0 → x = 3.
Domain {x ∈ ℝ | x > 3}; range {y ∈ ℝ}.
x-intercept: set y = 0. log (x − 3) = −2, so x − 3 = 10⁻² = 0.01 and x = 3.01.

A log curve through (3.1, 1), (4, 2) and (13, 3), hugging the asymptote x = 3; domain x > 3; range all reals; x-intercept 3.01.

Worked example

Example 2: Reflection and stretch

Describe the transformations and sketch y = −2 log x.

a = −2: a vertical stretch by a factor of 2 and a reflection in the x-axis.
Mapping rule: (x, y) → (x, −2y).
(0.1, −1) → (0.1, 2); (1, 0) → (1, 0); (10, 1) → (10, −2).
The asymptote stays x = 0. The graph is decreasing.

A decreasing log curve through (0.1, 2), (1, 0) and (10, −2) with asymptote x = 0.

Worked example

Example 3: All three parameters

For y = 3 log (x + 2) − 1: (a) describe the transformations, (b) find the images of the key points, (c) state the domain and asymptote, (d) find the intercepts.

(a) a = 3: vertical stretch by 3. d = −2: shift left 2. c = −1: shift down 1.
(b) Mapping rule (x, y) → (x − 2, 3y − 1): (0.1, −1) → (−1.9, −4); (1, 0) → (−1, −1); (10, 1) → (8, 2).
(c) Asymptote x = −2; domain {x ∈ ℝ | x > −2}; range {y ∈ ℝ}.
(d) y-intercept: x = 0 gives y = 3 log 2 − 1 ≈ −0.097.
(d) x-intercept: 3 log (x + 2) − 1 = 0, so log (x + 2) = 1/3, x + 2 = 10^(1/3) and x = 10^(1/3) − 2 ≈ 0.154.

Key points (−1.9, −4), (−1, −1), (8, 2); asymptote x = −2; y-intercept ≈ −0.097; x-intercept ≈ 0.154.

Worked example

Example 4: Equation from a description

The graph of y = log x is stretched vertically by a factor of 2, then translated 4 units left and 1 unit down. Write its equation and find the image of (10, 1).

a = 2, d = −4, c = −1.
y = 2 log (x + 4) − 1.
Image of (10, 1): (10 − 4, 2(1) − 1) = (6, 1). Check: 2 log 10 − 1 = 2 − 1 = 1 ✓

y = 2 log (x + 4) − 1, and (10, 1) maps to (6, 1).

Just for fun

When you translate a log graph, its asymptote moves with it. Asymptotes are the most loyal friends in mathematics: always close by, never touching, never in the way.

Hands-on activity

Make a log template

You need: a piece of thin cardboard (a cereal box panel works), scissors (cut slowly, keeping your fingers away from the blades), graph paper, a pencil and a calculator.

On graph paper with 1 cm per unit, draw y = log x carefully: plot (0.1, −1), (0.5, −0.30), (1, 0), (2, 0.30), (5, 0.70), (10, 1), and sketch the curve between them. Mark the asymptote x = 0.
Trace the curve onto the cardboard, along with a short line for the asymptote and a dot at (1, 0). Cut out a thin strip that follows the curve.
On a new set of axes, slide the template to draw y = log (x − 2), y = log x − 3 and y = log (x + 1) + 2. Keep the template upright (no flipping or stretching).
Which of the three graphs pass through the point (3, 0)?

What to look for: sliding the template only translates the graph; the dot (1, 0) lands at (3, 0), (1, −3) and (0, 2) respectively, and the asymptote lands at x = 2, x = 0 and x = −1. Only y = log (x − 2) passes through (3, 0). A template cannot show a stretch: that needs a new drawing, which is why a has to be applied to the points first.

Your turn

Describe the transformation that maps y = log x to y = log (x + 5).

Describe the transformation that maps y = log x to y = log x − 4.

State the domain and asymptote of y = log (x − 6) + 1.

Find the images of (0.1, −1), (1, 0) and (10, 1) under y = 4 log x.

Find the images of the same three points under y = −log (x − 1).

Find the x-intercept of y = log (x − 2) + 1.

Write the equation after y = log x is reflected in the x-axis and translated 3 units up.

More practice: warm-up

In y = log (x − d) + c, which parameter moves the asymptote?

Which way does y = log (x + 2) shift compared with y = log x?

What is the asymptote of y = log x + 7?

Is y = −3 log x increasing or decreasing?

Where does (1, 0) go under y = log (x − 4)?

More practice: on your own

Sketch y = (1/2) log (x + 3) by finding the images of the key points and the asymptote.

For y = 2 log (x − 1) + 3, find the y-intercept, if any.

For y = 2 log (x − 1) + 3, find the x-intercept.

A transformed log graph has asymptote x = −4 and passes through (−3, 2) and (6, 5). Find its equation in the form y = a log (x − d) + c.

Explain why y = log (x − 2) and y = log x − 2 are different functions, using one point.

Challenge

Find the equation y = a log x + c whose graph passes through (1, 5) and (100, −1).

For y = −2 log (x + 1) + 4, find where the graph crosses the line y = 6.

Show that y = log (100x) is a vertical translation of y = log x, and say how far.

Check your understanding

Quiz

How does the graph of y = log (x − 7) compare with y = log x?

What is the asymptote of y = log (x + 4) − 2?

Under y = 5 log x, the point (10, 1) maps to

What is the domain of y = 2 log (x − 3) + 1?

Which equation describes y = log x reflected in the x-axis and moved 1 unit left?

What is the x-intercept of y = log x + 1?

For parents and teachers

Big idea (A2.3): y = a log (x − d) + c is y = log x stretched by a (reflected if a < 0), moved d right and c up. The mapping rule (x, y) → (x + d, ay + c) applied to (0.1, −1), (1, 0), (10, 1) and the asymptote gives a quick, accurate sketch.

How to help: Ask your teen to say each transformation in words before sketching ("stretch by 3, left 2, down 1"), and then to name the new asymptote. If you have a computer, the free Desmos graphing calculator with sliders makes the effect of each parameter very clear.

Common difficulties:

Shifting the wrong way: log (x + 2) moves left, not right.
Forgetting that the asymptote moves with d, but not with c or a.
Applying the stretch after the translation (stretch or reflect first, then translate).
Trying to find a y-intercept when x = 0 is outside the domain.