2.3 Function Graphs, Transformations, and Zeros

2.3 Function Graphs, Transformations, and Zeros

1. Fundamental Graphical Transformations

Graph transformation allows the construction of complex function graphs from standard elementary base graphs ($y = x^2, y = e^x, y = \ln x, y = \sin x$).

A. Translations (平移变换)

• Horizontal Shift: $y = f(x) \to y = f(x \pm a)$ ($a > 0$). Shift left by $a$ units for $+a$, shift right by $a$ units for $-a$. ("左加右减")
• Vertical Shift: $y = f(x) \to y = f(x) \pm b$ ($b > 0$). Shift up by $b$ units for $+b$, shift down by $b$ units for $-b$. ("上加下减")

Caution on Horizontal Scaling/Translation Order: When transforming $y = f(x)$ to $y = f(\omega x + \phi)$, factor out \omega: $y = f\left(\omega\left(x + \frac{\phi}{\omega}\right)\right)$. The horizontal shift is $\frac{\phi}{\omega}$ units, NOT $\phi$ units.

B. Scaling / Stretching (伸缩变换)

• Vertical Scaling: $y = A f(x)$ ($A > 0$). Stretch vertically by factor $A$ if $A > 1$; compress by factor $A$ if $0 < A < 1$.
• Horizontal Scaling: $y = f(\omega x)$ (\omega > 0). Compress horizontally by factor $\frac{1}{\omega}$ if $\omega > 1$; stretch horizontally by factor $\frac{1}{\omega}$ if $0 < \omega < 1$.

C. Reflection Transformations (对称变换)

• $y = -f(x)$: Reflect $y = f(x)$ across the $x$-axis.
• $y = f(-x)$: Reflect $y = f(x)$ across the $y$-axis.
• $y = -f(-x)$: Reflect $y = f(x)$ across the origin $(0,0)$.
• $y = f^{-1}(x)$: Reflect $y = f(x)$ across the line $y = x$ (inverse function graph).

D. Absolute Value Transformations (翻折变换)

1. $y = |f(x)|$: Retain the graph of $f(x)$ where $f(x) \ge 0$ (above or on the $x$-axis). Reflect the portion where $f(x) < 0$ (below the $x$-axis) upward across the $x$-axis.
2. $y = f(|x|)$: Retain the graph of $f(x)$ for $x \ge 0$ (on and to the right of the $y$-axis). Discard the graph for $x < 0$, and reflect the right-hand graph across the $y$-axis to form the left half.

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2. Four-Step Graph Identification Strategy for Gaokao

Multiple-choice graph identification questions test structural recognition. Apply this 4-step elimination process:

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3. Zeros of Functions and Root Existence Theorem

A. Concept of Function Zeros

A real number $x_0$ is called a zero of the function $f(x)$ if $f(x_0) = 0$. The following three statements are mathematically equivalent:

1. $x_0$ is a zero of the function $y = f(x)$.
2. $x_0$ is a real root of the equation $f(x) = 0$.
3. $x_0$ is the $x$-intercept of the graph of $y = f(x)$ (where the graph intersects the line $y = 0$).

B. Bolzano's Zero Existence Theorem (零点存在性定理)

If $y = f(x)$ is continuous on a closed interval $[a, b]$, and $f(a) \cdot f(b) < 0$ (opposite signs at endpoints), then there exists at least one $c \in (a, b)$ such that $f(c) = 0$.

Uniqueness Condition: If $f(x)$ is continuous on $[a, b]$, $f(a) \cdot f(b) < 0$, and $f(x)$ is strictly monotonic on $[a, b]$, then $f(x)$ has exactly one unique zero $c \in (a, b)$.

C. Parameter Isolation Method (分离参数法)

When finding parameter ranges $k$ such that $f(x, k) = 0$ has solutions on an interval:

1. Rearrange the equation into isolated form $k = g(x)$.
2. The problem reduces to finding the range of $g(x)$, or determining the number of intersection points between the horizontal line $y = k$ and the curve $y = g(x)$.

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4. Worked Gaokao Exam Examples

Example 1: Determining Parameter Range for Multiple Zeros

Problem: Find the range of the parameter $a$ such that the function $f(x) = |x^2 - 2x| - a$ has exactly $3$ distinct real zeros.

Solution:

1. Convert zero condition to graph intersection: $f(x) = 0 \iff |x^2 - 2x| = a$. This is equivalent to the number of intersection points between $y = g(x) = |x^2 - 2x|$ and $y = a$.
2. Analyze the base quadratic $h(x) = x^2 - 2x = (x-1)^2 - 1$:
   • Roots of $h(x) = 0$ are $x = 0$ and $x = 2$. Vertex is at $(1, -1)$.
3. Apply absolute value transformation $g(x) = |h(x)|$:
   • For $x \le 0$ or $x \ge 2$, $g(x) = x^2 - 2x$.
   • For $0 < x < 2$, $g(x) = -(x^2 - 2x) = 2x - x^2$. The local maximum is at $(1, 1)$.
4. Analyze intersections with horizontal line $y = a$:
   • If $a < 0$: $0$ intersections.
   • If $a = 0$: $2$ intersections at $(0,0)$ and $(2,0)$.
   • If $0 < a < 1$: $4$ distinct intersections.
   • If $a = 1$: $3$ distinct intersections at $x = 1, 1 - \sqrt{2}, 1 + \sqrt{2}$.
   • If $a > 1$: $2$ distinct intersections. Conclusion: The range of $a$ for exactly $3$ distinct zeros is $a = 1$.

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Example 2: Zero Existence Interval and Monotonicity

Problem: Prove that $f(x) = \ln x + 2x - 6$ has a unique zero in the interval $(2, 3)$.

Solution:

1. Domain check: $D_f = (0, +\infty)$. Function $f(x)$ is continuous on $(0, +\infty)$.
2. Monotonicity check:
   • Derivative $f'(x) = \frac{1}{x} + 2$.
   • For all $x > 0$, $\frac{1}{x} > 0$ and $2 > 0$, so $f'(x) > 0$.
   • Thus $f(x)$ is strictly increasing on $(0, +\infty)$.
3. Endpoint sign test on $[2, 3]$:
   • $f(2) = \ln 2 + 2(2) - 6 = \ln 2 - 2$. Since $e^2 \approx 7.389 > 2$, $\ln 2 < 2 \implies f(2) < 0$.
   • $f(3) = \ln 3 + 2(3) - 6 = \ln 3 > 0$.
4. Apply Bolzano's Theorem & Uniqueness:
   • $f(2) \cdot f(3) < 0$ and $f(x)$ is strictly increasing on $[2, 3]$.
   • Therefore, $f(x)$ has a unique zero in $(2, 3)$.