Unit 3 · Elementary Functions　基本初等函数

By the end of this chapter you can:

1. Recognise and sketch power, exponential, logarithmic, and quadratic functions and read off their key features
2. Apply log laws and the change-of-base formula to simplify expressions without a calculator
3. Convert a quadratic to vertex form, locate the vertex, and use the discriminant to count roots
4. Apply translation, scaling, and reflection rules to shift any curve in the plane

Exam weight on past CSCA papers: ~10% (4–5 of 48 MCQs).

3.1　Power Functions　幂函数

A power function has the form
$$f(x) = x^{\alpha}$$
where $\alpha$ is a fixed real number (the exponent). The base is always $x$; only the exponent is the "name" of the function.

Five canonical power functions

Three things every power function graph passes through (when defined there): $(0,0)$, $(1,1)$, and for odd $\alpha$ also $(-1,-1)$. These are pinch points you can always verify on exam day.

Symmetry pattern

• Even exponent ($\alpha = 2, 4, \ldots$): $f(-x) = f(x)$ → graph is symmetric about the $y$-axis.
• Odd exponent ($\alpha = 1, 3, \ldots$): $f(-x) = -f(x)$ → graph is symmetric about the origin.
• Fractional/negative exponent: check case by case (domain may not be symmetric).

🔑 Exam fast-track: the CSCA often asks "which graph is a power function?" or "which power function is increasing on all of $\mathbb{R}$?" — $y = x^{3}$ is the only one among the five canonical cases that is strictly increasing on all of $\mathbb{R}$ and has point symmetry about the origin.

Growth comparison for $x > 1$ vs $0 < x < 1$

For $x > 1$ and positive exponents: larger exponent → faster growth. So $x^{3} > x^{2} > x$ when $x > 1$.

⚠️ Trap: the order reverses on $(0,1)$. If $0 < x < 1$, then $x^{3} < x^{2} < x$ — multiplying a proper fraction by itself makes it smaller. This is a favourite MCQ trap.

Worked Example 3.1.A

Which of the following is a power function that is decreasing on $(0, +\infty)$?

(A) $y = x^{2}$   (B) $y = x^{-1}$   (C) $y = x^{3}$   (D) $y = x^{1/2}$

Solution.

Recall that $f(x) = x^{\alpha}$ is decreasing on $(0,\infty)$ when $\alpha < 0$, because a larger denominator makes the fraction smaller.

• $y = x^{2}$: $\alpha = 2 > 0$ → increasing on $(0,\infty)$. ✗
• $y = x^{-1} = \dfrac{1}{x}$: $\alpha = -1 < 0$ → as $x$ increases, $\dfrac{1}{x}$ decreases. ✓
• $y = x^{3}$: $\alpha = 3 > 0$ → increasing. ✗
• $y = x^{1/2}$: $\alpha = \tfrac{1}{2} > 0$ → increasing. ✗

$$\boxed{B}$$

3.2　Exponential Functions　指数函数

An exponential function has the form
$$f(x) = a^{x}, \quad a > 0,\ a \ne 1.$$
The base $a$ is fixed; the variable $x$ is the exponent. This is the opposite arrangement from a power function — compare $x^{2}$ (power) with $2^{x}$ (exponential).

Key properties

⚠️ Trap: $a^{x}$ is never zero and never negative — its range is strictly $(0, +\infty)$. A common MCQ distractor offers $(-\infty, +\infty)$ or $[0, +\infty)$ as the range; both are wrong.

The natural exponential $e^{x}$

When $a = e \approx 2.718$, we write $f(x) = e^{x}$. Since $e > 1$, it is increasing. You will also encounter $e^{x}$ in Unit 3 via natural logs.

🔑 Symmetry trick: the graphs of $y = a^{x}$ and $y = \left(\tfrac{1}{a}\right)^{x} = a^{-x}$ are reflections of each other in the $y$-axis. If you know one, flip it horizontally to get the other.

Worked Example 3.2.A

Without a calculator, determine which is larger: $0.3^{-2}$ or $0.3^{-3}$.

Solution.

Let $f(x) = 0.3^{x}$. Base $0.3 < 1$, so $f$ is decreasing in $x$.

Since $-2 > -3$, and $f$ is decreasing, we get $f(-2) < f(-3)$, i.e.
$$0.3^{-2} < 0.3^{-3}.$$

Sanity check: $0.3^{-2} = \tfrac{1}{0.09} \approx 11.1$ and $0.3^{-3} = \tfrac{1}{0.027} \approx 37.0$. ✓

$$\boxed{0.3^{-2} < 0.3^{-3}}$$

3.3　Logarithmic Functions　对数函数

The logarithmic function is the inverse of the exponential:
$$y = \log_{a} x \iff a^{y} = x, \quad a > 0,\ a \ne 1,\ x > 0.$$

Read "$\log_{a} x$" as "the power you raise $a$ to in order to get $x$."

Key properties

The domain is $(0, +\infty)$: you can never take the logarithm of zero or a negative number.

The three log laws　对数运算法则

$$\log_{a}(MN) = \log_{a} M + \log_{a} N \tag{product rule}$$
$$\log_{a}\!\left(\frac{M}{N}\right) = \log_{a} M - \log_{a} N \tag{quotient rule}$$
$$\log_{a}(M^{p}) = p\,\log_{a} M \tag{power rule}$$

Change-of-base formula　换底公式

$$\log_{a} b = \frac{\log_{c} b}{\log_{c} a} \quad \text{(any valid base } c > 0,\ c \ne 1).$$

Most useful forms: $\log_{a} b = \dfrac{\ln b}{\ln a} = \dfrac{\lg b}{\lg a}$.

Reciprocal identity: $\log_{a} b = \dfrac{1}{\log_{b} a}$.

🔑 Sign of a log at a glance:

• $\log_{a} x > 0$ when $a > 1, x > 1$ OR when $0 < a < 1, 0 < x < 1$
• $\log_{a} x < 0$ when $a > 1, 0 < x < 1$ OR when $0 < a < 1, x > 1$
• $\log_{a} 1 = 0$ always (since $a^{0} = 1$ for any valid base).

⚠️ Trap: there is no log law for a sum. $\log_{a}(M + N) \ne \log_{a} M + \log_{a} N$. This is one of the most common errors on CSCA.

Worked Example 3.3.A

Simplify without a calculator: $\dfrac{\log_{4} 32}{\log_{4} 2}$.

Solution. Recognise the pattern: $\dfrac{\log_{a} M}{\log_{a} N} = \log_{N} M$ (change-of-base in reverse).

$$\frac{\log_{4} 32}{\log_{4} 2} = \log_{2} 32 = \log_{2} 2^{5} = 5.$$

Verification using direct computation: $32 = 2^{5}$, so $\log_{4} 32 = \log_{4} 4^{5/2} = \tfrac{5}{2}$; and $\log_{4} 2 = \log_{4} 4^{1/2} = \tfrac{1}{2}$. Ratio: $\dfrac{5/2}{1/2} = 5$. ✓

$$\boxed{5}$$

3.4　Quadratic Functions　二次函数

A quadratic function has the form
$$f(x) = ax^{2} + bx + c, \quad a \ne 0.$$

Three equivalent forms

Completing the square converts standard form to vertex form:
$$ax^{2} + bx + c = a\!\left(x + \frac{b}{2a}\right)^{2} + c - \frac{b^{2}}{4a}.$$

The vertex is therefore $\left(-\dfrac{b}{2a},\ c - \dfrac{b^{2}}{4a}\right)$.

The discriminant　判别式

$$\Delta = b^{2} - 4ac.$$

Opening direction and extrema

🔑 "a positive, arms up; a negative, arms down."

• $a > 0$: parabola opens upward — vertex is a minimum.
• $a < 0$: parabola opens downward — vertex is a maximum.

The axis of symmetry is the vertical line $x = -\dfrac{b}{2a}$.

⚠️ Trap: "maximum value" and "maximum $x$-value" are different. The maximum value of $f$ is the $y$-coordinate of the vertex, $k = f\!\left(-\tfrac{b}{2a}\right)$. Many students write the $x$-coordinate instead.

Worked Example 3.4.A

The quadratic $f(x) = -2x^{2} + 8x - 5$ — find its vertex and maximum value.

Solution.

$a = -2$, $b = 8$, $c = -5$.

Axis: $x = -\dfrac{8}{2(-2)} = 2$.

Maximum value:
$$f(2) = -2(4) + 8(2) - 5 = -8 + 16 - 5 = 3.$$

Using the vertex formula directly: $k = c - \dfrac{b^{2}}{4a} = -5 - \dfrac{64}{4(-2)} = -5 + 8 = 3$. ✓

Since $a = -2 < 0$, the parabola opens downward, confirming this is indeed a maximum.

$$\boxed{\text{Vertex } (2,\ 3),\quad \text{maximum value } = 3.}$$

3.5　Function Translations　函数图像变换

Given any "base" graph $y = f(x)$, you can produce new graphs by four types of transformation. These rules apply uniformly to every function type in this unit — power, exponential, logarithmic, quadratic, or anything else.

The transformation rules

⚠️ Trap — horizontal shifts are counterintuitive. $y = f(x - 3)$ shifts the graph right by 3, not left. Reason: to produce the same output, $x$ must be 3 larger than before, so every point moves right. Substitute $x = 3$ and you get the original $f(0)$ — the graph anchors at a point further right.

🔑 Order when multiple transformations combine: work from the inside out.

1. Horizontal shift (inside the argument)
2. Horizontal scale/reflection (inside the argument)
3. Vertical scale/reflection (outside)
4. Vertical shift (outermost)

Symmetry via reflections

• $y = f(-x)$ is the reflection of $y = f(x)$ in the $y$-axis.
• $y = -f(x)$ is the reflection in the $x$-axis.
• $y = -f(-x)$ is the point reflection through the origin (equivalent to 180° rotation).

Worked Example 3.5.A

Starting from $y = \log_{2} x$, describe the steps needed to obtain $y = \log_{2}(x + 3) - 1$, and state the new vertical asymptote and $x$-intercept.

Solution.

Identify changes:

• $x$ replaced by $x + 3$ (inside the log) → shift left by 3.
• $-1$ added outside → shift down by 1.

Original vertical asymptote: $x = 0$. After shifting left by 3: $x = -3$.

$x$-intercept (set $y = 0$):
$$\log_{2}(x + 3) - 1 = 0 \implies \log_{2}(x+3) = 1 \implies x + 3 = 2^{1} = 2 \implies x = -1.$$

$$\boxed{\text{Vertical asymptote: } x = -3;\quad x\text{-intercept: } (-1,\ 0).}$$

Check: the original passes through $(1, 0)$. Shifting left by 3 moves this to $(-2, 0)$; shifting down by 1 moves it to $(-2, -1)$. The new $x$-intercept is at $x = -1$ (where $y = 0$), consistent with our calculation. ✓

Try it!　自测练习

Q1. On the interval $(0, 1)$, which is larger: $x^{2}$ or $x^{3}$? Justify your answer.

Q2. The graph of $y = a^{x}$ passes through the point $(2, 9)$. Find $a$ and state whether $f(x) = a^{x}$ is increasing or decreasing.

Q3. Simplify without a calculator: $\log_{2} 6 - \log_{2} 3 + \log_{2} 8$.

Q4. Find the vertex of $f(x) = 2x^{2} - 4x + 5$ and state whether it is a maximum or minimum.

Q5. The graph of $y = \sqrt{x}$ is reflected in the $y$-axis and then shifted up by 2. Write the equation of the resulting curve and state its domain.

📌 Chapter summary

Topic	Key skill
Power functions $x^{\alpha}$	5 canonical cases; order reverses on $(0,1)$; even/odd symmetry by exponent parity
Exponential $a^{x}$	$a>1$: increasing; $0<a<1$: decreasing; range always $(0,+\infty)$; asymptote $y=0$
Logarithmic $\log_{a}x$	Inverse of $a^{x}$; domain $(0,+\infty)$; 3 log laws + change-of-base; $\log_{a}1=0$ always
Quadratic $ax^{2}+bx+c$	Vertex at $x=-b/2a$; discriminant $\Delta=b^{2}-4ac$ counts real roots; $a>0$→min, $a<0$→max
Graph transformations	$f(x-h)$: right $h$; $f(x)+k$: up $k$; $-f(x)$: flip over $x$-axis; $f(-x)$: flip over $y$-axis

What's next → Unit 4 applies the graph-transformation skills from §3.5 directly to trigonometric functions: $y = A\sin(\omega x + \phi) + k$ is exactly the combination of vertical stretch, horizontal compression, horizontal shift, and vertical shift you practised here.