二次函数èrcì hánshù

Core Concept

A quadratic function is a polynomial function of degree 2:

$f(x) = ax^2 + bx + c \quad (a \neq 0)$

where $a$, $b$, $c$ are constants and $a \neq 0$.

Three Forms

Form	Expression	Key Feature
Standard form	$y = ax^2 + bx + c$	Shows y-intercept $c$
Vertex form	$y = a(x-h)^2 + k$	Shows vertex $(h, k)$
Factored form	$y = a(x-x_1)(x-x_2)$	Shows x-intercepts $x_1, x_2$

The Vertex

The vertex is the highest or lowest point of the parabola.

Vertex Coordinates

$h = -\frac{b}{2a}, \quad k = \frac{4ac - b^2}{4a}$

Or equivalently: $k = f(h) = f\left(-\frac{b}{2a}\right)$

Vertex Form

$f(x) = a(x - h)^2 + k$

where $(h, k)$ is the vertex.

Graph Properties

Direction of Opening

• $a > 0$: Parabola opens upward (U-shape), vertex is minimum
• $a < 0$: Parabola opens downward (∩-shape), vertex is maximum

Axis of Symmetry

$x = -\frac{b}{2a} = h$

The parabola is symmetric about this vertical line.

Y-intercept

The y-intercept is at $(0, c)$, found by setting $x = 0$.

X-intercepts (Roots)

Found by solving $ax^2 + bx + c = 0$. The discriminant $\Delta = b^2 - 4ac$ determines:

• $\Delta > 0$: Two distinct x-intercepts
• $\Delta = 0$: One x-intercept (vertex touches x-axis)
• $\Delta < 0$: No x-intercepts

Monotonicity

When $a > 0$:

• Decreasing on $(-\infty, h]$
• Increasing on $[h, +\infty)$

When $a < 0$:

• Increasing on $(-\infty, h]$
• Decreasing on $[h, +\infty)$

Range

When $a > 0$:

$\text{Range} = [k, +\infty)$

When $a < 0$:

$\text{Range} = (-\infty, k]$

CSCA Practice Problems

💡 Note: The following practice problems are designed based on the CSCA exam syllabus.

Example 1: Basic (Difficulty ★★☆☆☆)

Find the vertex of $f(x) = x^2 - 6x + 5$.

Solution:

Method 1 (Formula): $h = -\frac{-6}{2(1)} = 3$ $k = f(3) = 9 - 18 + 5 = -4$

Method 2 (Completing the square): $f(x) = (x^2 - 6x + 9) - 9 + 5 = (x-3)^2 - 4$

Answer: Vertex is $(3, -4)$

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Example 2: Intermediate (Difficulty ★★★☆☆)

Find the range of $f(x) = -x^2 + 4x - 1$ on $[0, 3]$.

Solution:

Complete the square: $f(x) = -(x^2 - 4x) - 1 = -(x-2)^2 + 4 - 1 = -(x-2)^2 + 3$

Vertex at $(2, 3)$, parabola opens downward.

Since $2 \in [0, 3]$, maximum is at vertex: $f(2) = 3$

Check endpoints:

• $f(0) = -1$
• $f(3) = -9 + 12 - 1 = 2$

Minimum is $f(0) = -1$.

Answer: Range is $[-1, 3]$

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Example 3: Advanced (Difficulty ★★★★☆)

If $f(x) = x^2 - 2ax + a$ has minimum value $-2$ on $[0, 2]$, find $a$.

Solution:

$f(x) = (x-a)^2 + a - a^2$, vertex at $(a, a-a^2)$.

Case 1: $a < 0$ (vertex left of interval) Minimum at $x = 0$: $f(0) = a = -2$ Check: vertex at $(-2, -2-4) = (-2, -6)$, but $f(0) = -2 \neq -6$. ✓ So $a = -2$ works.

Case 2: $0 \leq a \leq 2$ (vertex inside interval) Minimum at vertex: $a - a^2 = -2$ $a^2 - a - 2 = 0$ $(a-2)(a+1) = 0$ $a = 2$ or $a = -1$ Only $a = 2$ is in $[0, 2]$. Check: $f(x) = (x-2)^2$, min at $x=2$ is $0 \neq -2$. ✗

Case 3: $a > 2$ (vertex right of interval) Minimum at $x = 2$: $f(2) = 4 - 4a + a = 4 - 3a = -2$ $a = 2$, contradicts $a > 2$. ✗

Answer: $a = -2$

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Example 4: Advanced (Difficulty ★★★★☆)

Find all values of $m$ such that $f(x) = x^2 - mx + 1 > 0$ for all $x \in \mathbb{R}$.

Solution:

For $f(x) > 0$ for all $x$, the parabola must open upward (✓, $a = 1 > 0$) and have no x-intercepts.

This requires $\Delta < 0$: $\Delta = m^2 - 4(1)(1) = m^2 - 4 < 0$ $m^2 < 4$ $-2 < m < 2$

Answer: $m \in (-2, 2)$

Converting Between Forms

Standard to Vertex Form

Given $f(x) = ax^2 + bx + c$:

1. Factor out $a$ from first two terms: $f(x) = a(x^2 + \frac{b}{a}x) + c$
2. Complete the square: $f(x) = a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c$
3. Simplify: $f(x) = a(x - h)^2 + k$ where $h = -\frac{b}{2a}$, $k = c - \frac{b^2}{4a}$

Vertex to Standard Form

Given $f(x) = a(x-h)^2 + k$:

Expand: $f(x) = ax^2 - 2ahx + ah^2 + k$

So $b = -2ah$ and $c = ah^2 + k$.

Common Mistakes

❌ Mistake 1: Wrong Sign in Vertex Formula

Wrong: $h = \frac{b}{2a}$ ✗

Correct: $h = -\frac{b}{2a}$ ✓

❌ Mistake 2: Ignoring Domain When Finding Range

Wrong: Range of $f(x) = x^2$ on $[1, 3]$ is $[0, +\infty)$ ✗

Correct: On $[1, 3]$, minimum is $f(1) = 1$, so range is $[1, 9]$ ✓

❌ Mistake 3: Confusing Vertex Location

When domain is restricted, the vertex might be outside the domain. Check if vertex is inside, left, or right of the interval.

Study Tips

1. ✅ Master completing the square: Essential for finding vertex
2. ✅ Know the three cases: Vertex inside, left of, or right of interval
3. ✅ Use discriminant: For questions about intersections with x-axis
4. ✅ Draw sketches: Visualize parabola to avoid errors

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💡 Exam Tip: For restricted domain problems, first locate the vertex relative to the domain, then check endpoints. The extreme values occur either at the vertex (if inside domain) or at endpoints!