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Quadratic Functions

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Quadratic Functions

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Quadratic Functions

🎯 Learning Objectives

By the end of this topic, you will be able to:

Identify a quadratic function from its standard form $f(x) = ax^2 + bx + c$
Find the vertex, axis of symmetry, and intercepts of a parabola
Determine whether the parabola opens upward or downward
Sketch the graph of a quadratic function
Find the maximum or minimum value of a quadratic function
Compute the slope of a quadratic function at a point

📋 Prerequisites

Straight Lines — slope, intercepts, linear equations
Functions — domain, range, function notation
Algebra — completing the square, solving linear equations

📖 Core Content

3.1 Intuition: The Shape of a Ball in Flight

Throw a ball into the air. It rises, slows, reaches a peak, then falls back down. If you plot its height over time, you get a smooth U-shaped curve called a parabola. This is the graph of a quadratic function. Quadratic functions model:

Projectile motion (height vs. time)
Profit optimization (revenue vs. price)
Area problems (maximizing space with fixed perimeter)
The shape of satellite dishes and headlight reflectors

💡 Why this matters: Quadratics are the simplest non-linear functions. Understanding them is the first step toward understanding all higher-degree polynomials and the curvature of functions — essential for optimization in machine learning (gradient descent finds minima of functions).

3.2 Formal Definition

A quadratic function is any function that can be written in the form:

f(x) = ax^2 + bx + c

where

a

b

c

are real constants and

a \neq 0

(if

a = 0

, it's linear).

Component	Name	Role
$a$	Leading coefficient	Determines "width" and direction (up/down) of parabola
$b$	Linear coefficient	Affects position of vertex horizontally
$c$	Constant term	$y$ -intercept: $f(0) = c$
$ax^2$	Quadratic term	Gives the U-shape

(Diagram)

3.3 Key Features of a Parabola

3.3.1 Vertex

The vertex is the turning point of the parabola — where it changes from increasing to decreasing (or vice versa). It is the minimum (if

a > 0

) or maximum (if

a < 0

x_{\text{vertex}} = -\frac{b}{2a}

y_{\text{vertex}} = f\left(-\frac{b}{2a}\right) = c - \frac{b^2}{4a}

Derivation (completing the square): f(x) &= ax^2 + bx + c \\[2pt] &= a\left(x^2 + \frac{b}{a}x\right) + c \\[2pt] &= a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\right] + c \\[2pt] &= a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right) \end{aligned}

This is the **vertex form**: $f(x) = a(x - h)^2 + k$, where $(h, k)$ is the vertex. h = -\\frac{b}{2a},\\quad k = c - \\frac{b^2}{4a} \#### 3.3.2 Axis of Symmetry The vertical line through the vertex about which the parabola is symmetric: x = -\\frac{b}{2a} \#### 3.3.3 Intercepts - **$y$-intercept:** $f(0) = c$ → point $(0, c)$ - **$x$-intercepts:** Solve $ax^2 + bx + c = 0$ (covered in detail in Quadratics — Equations) #### 3.3.4 Direction and Width | $a$ | Direction | Shape | |:---:|:---|:---| | $a > 0$ | Opens upward (U) | Minimum at vertex | | $a < 0$ | Opens downward (∩) | Maximum at vertex | | Larger $|a|$ | Narrower parabola | Steeper sides | | Smaller $|a|$ | Wider parabola | Flatter sides | --- ### 3.4 Graphing a Quadratic Function **Recipe for graphing $f(x) = ax^2 + bx + c$:** 1. Determine direction: $a > 0$ (up) or $a < 0$ (down) 2. Find vertex: $x_v = -b/(2a)$, $y_v = f(x_v)$ 3. Find $y$-intercept: $(0, c)$ 4. Find $x$-intercepts (if any) by solving $ax^2 + bx + c = 0$ 5. Plot additional points if needed (use symmetry) 6. Draw the parabola through these points ### 3.5 Worked Examples **Example 1.1 (Easy):** For $f(x) = x^2 - 4x + 3$, find the vertex, axis of symmetry, intercepts, and graph. *Step 1 — Direction:* $a = 1 > 0$ → opens upward *Step 2 — Vertex:* $x_v = -\\frac{-4}{2(1)} = 2$, $y_v = f(2) = 4 - 8 + 3 = -1$ → Vertex: $(2, -1)$ *Step 3 — Axis of symmetry:* $x = 2$ *Step 4 — y-intercept:* $(0, 3)$ *Step 5 — x-intercepts:* $x^2 - 4x + 3 = 0 \\implies (x-1)(x-3) = 0 \\implies x = 1, 3$ Points: $(1, 0)$, $(3, 0)$ \\boxed{\\text{Vertex }(2,-1),\\ \\text{Axis }x=2,\\ \\text{y-int }(0,3),\\ \\text{x-ints }(1,0),(3,0)} **Example 1.2 (Medium):** For $f(x) = -2x^2 + 8x - 5$, find the vertex and state the maximum value. *Step 1 — Direction:* $a = -2 < 0$ → opens downward → has a maximum *Step 2 — Vertex:* $x_v = -\\frac{8}{2(-2)} = -\\frac{8}{-4} = 2$ $y_v = f(2) = -2(4) + 8(2) - 5 = -8 + 16 - 5 = 3$ Vertex: $(2, 3)$ *Step 3 — Maximum value:* $3$ (occurs at $x = 2$) \\boxed{\\text{Maximum } = 3 \\text{ at } x = 2} **Example 1.3 (Hard):** A ball is thrown upward with height $h(t) = -5t^2 + 20t + 2$ meters after $t$ seconds. Find: a) The maximum height b) When it hits the ground c) The initial height *Step 1 — Vertex (maximum):* $t_v = -\\frac{20}{2(-5)} = 2$ seconds $h(2) = -5(4) + 20(2) + 2 = -20 + 40 + 2 = 22$ m *Step 2 — Hits ground ($h=0$):* $-5t^2 + 20t + 2 = 0$ Multiply by $-1$: $5t^2 - 20t - 2 = 0$ Using quadratic formula: $t = \\frac{20 \\pm \\sqrt{400 + 40}}{10} = \\frac{20 \\pm \\sqrt{440}}{10} = \\frac{20 \\pm 2\\sqrt{110}}{10} = 2 \\pm \\frac{\\sqrt{110}}{5}$ Positive root: $t \\approx 2 + 2.1 = 4.1$ seconds *Step 3 — Initial height:* $h(0) = 2$ m \\boxed{\\text{Max height } 22\\text{m at }t=2\\text{s},\\ \\text{lands at }t\\approx4.1\\text{s},\\ \\text{initial }2\\text{m}} \--- ### 3.6 Slope of a Quadratic Function The slope of a quadratic changes at every point (unlike a line which has constant slope). We'll study this in detail in calculus (Week 8), but for now: At any point $x$, the slope of $f(x) = ax^2 + bx + c$ is: m(x) = 2ax + b \- At the vertex $x = -b/(2a)$, the slope is $0$ (horizontal tangent) - For $x < -b/(2a)$, the slope is negative (if $a > 0$) or positive (if $a < 0$) - For $x > -b/(2a)$, the slope is positive (if $a > 0$) or negative (if $a < 0$) **Example:** Find the slope of $f(x) = x^2 - 4x + 3$ at $x = 1$ and $x = 3$. $m(x) = 2x - 4$ At $x = 1$: $m(1) = 2 - 4 = -2$ (decreasing) At $x = 3$: $m(3) = 6 - 4 = 2$ (increasing) --- ### 3.7 Vertex Form Transformations The vertex form $f(x) = a(x - h)^2 + k$ reveals the **transformations** of the basic parabola $y = x^2$: | Transformation | Effect | Equation | |:---|:---|:---| | Shift right $h$ | Vertex moves to $(h, k)$ | $y = (x - h)^2$ | | Shift left $h$ | Vertex moves to $(-h, k)$ | $y = (x + h)^2$ | | Shift up $k$ | Vertex moves to $(h, k)$ | $y = x^2 + k$ | | Shift down $k$ | Vertex moves to $(h, -k)$ | $y = x^2 - k$ | | Vertical stretch | Parabola narrows | $y = a x^2$, $a > 1$ | | Vertical compression | Parabola widens | $y = a x^2$, $0 < a < 1$ | | Reflection | Opens downward | $y = -x^2$ | --- ## 📐 Key Formulas — Summary Table | Concept | Formula | Notes | |:---|:---|:---| | Standard form | $f(x) = ax^2 + bx + c$ | $a \\neq 0$ | | Vertex form | $f(x) = a(x - h)^2 + k$ | $(h, k)$ is vertex | | Vertex $x$ | $h = -\\frac{b}{2a}$ | | | Vertex $y$ | $k = f(h) = c - \\frac{b^2}{4a}$ | | | Axis of symmetry | $x = -\\frac{b}{2a}$ | | | $y$-intercept | $(0, c)$ | | | Opening direction | Up if $a > 0$, down if $a < 0$ | | | Slope at $x$ | $m(x) = 2ax + b$ | | | Vertex from completing square | $a\\left(x + \\frac{b}{2a}\\right)^2 + \\left(c - \\frac{b^2}{4a}\\right)$ | | | Discriminant | $D = b^2 - 4ac$ | Nature of $x$-intercepts | --- ## ⚠️ Common Pitfalls ### Pitfall 1: Sign Error in Vertex Formula **Mistake:** Using $x_v = b/(2a)$ instead of $-b/(2a)$. **Why:** Forgetting the negative sign from the derivation. **Correct:** $x_v = -b/(2a)$. Always check: $f(x) = x^2$ has vertex at $(0,0)$, and $-b/(2a) = 0$ ✓. **Memory aid:** "The vertex is the OPPOSITE of $b$ over $2a$." ### Pitfall 2: Confusing $a$ and Its Effect on Shape **Mistake:** Thinking $a > 1$ makes the parabola wider. **Correct:** $|a| > 1$ makes the parabola **narrower** (steeper sides). $0 < |a| < 1$ makes it **wider**. **Example:** $y = 0.1x^2$ is much wider than $y = x^2$. ### Pitfall 3: Forgetting the $a$ When Completing the Square **Mistake:** $ax^2 + bx = a\\left(x^2 + \\frac{b}{a}x\\right)$ but forgetting to factor $a$ correctly. **Correct:** $3x^2 + 12x = 3(x^2 + 4x) = 3\[(x+2)^2 - 4\] = 3(x+2)^2 - 12$ ### Pitfall 4: Thinking All Parabolas Have Two $x$-intercepts **Correct:** A parabola may have 0, 1, or 2 $x$-intercepts depending on the discriminant. $y = x^2 + 1$ has no real $x$-intercepts (it sits entirely above the x-axis). --- ## 📝 Practice Questions > **Q1: Find the vertex of $f(x) = x^2 + 6x + 5$.** > > **Strategy Hint:** Use $x_v = -b/(2a)$. > > $x_v = -\\frac{6}{2(1)} = -3$ > $y_v = f(-3) = 9 - 18 + 5 = -4$ > >

\boxed{(-3, -4)}

> **Q2: Determine if $f(x) = -3x^2 + 12x - 7$ has a max or min, and find it.** > > **Strategy Hint:** $a < 0$ → max. Find vertex. > > $x_v = -\\frac{12}{2(-3)} = -\\frac{12}{-6} = 2$ > $y_v = -3(4) + 12(2) - 7 = -12 + 24 - 7 = 5$ > >

\boxed{\text{Maximum } 5 \text{ at } x = 2}

> **Q3: Find the axis of symmetry of $f(x) = 2x^2 - 8x + 1$.** > > **Strategy Hint:** $x = -b/(2a)$. > > $x = -\\frac{-8}{2(2)} = \\frac{8}{4} = 2$ > >

\boxed{x = 2}

> **Q4: Write $f(x) = x^2 - 10x + 28$ in vertex form.** > > **Strategy Hint:** Complete the square. > > $x^2 - 10x + 28 = (x^2 - 10x + 25) + 3 = (x - 5)^2 + 3$ > >

\boxed{f(x) = (x - 5)^2 + 3}

> **Q5: Find the $y$-intercept of $f(x) = -2x^2 + 3x + 4$.** > > **Strategy Hint:** Evaluate $f(0)$. > > $f(0) = 4$ > >

\boxed{(0, 4)}

> **Q6: The profit $P(x) = -x^2 + 50x - 200$ models profit from producing $x$ units. Find the production level that maximizes profit and the maximum profit.** > > **Strategy Hint:** Vertex gives max (since $a < 0$). > > $x_v = -\\frac{50}{2(-1)} = 25$ units > $P(25) = -(25)^2 + 50(25) - 200 = -625 + 1250 - 200 = 425$ > >

\boxed{25 \text{ units gives profit } \$425}

> **Q7: Find the slope of $f(x) = 3x^2 - 12x + 7$ at $x = 0$, $x = 2$, $x = 4$.** > > **Strategy Hint:** $m(x) = 2ax + b = 6x - 12$. > > $m(0) = -12$ > $m(2) = 0$ > $m(4) = 12$ > >

\boxed{-12,\ 0,\ 12}

> **Q8: Identify the transformations from $y = x^2$ to $y = -2(x + 3)^2 + 5$.** > > **Strategy Hint:** Compare to vertex form $a(x-h)^2 + k$. > > $a = -2$: reflection across $x$-axis + vertical stretch > $h = -3$: shift left 3 units > $k = 5$: shift up 5 units > >

\boxed{\text{Reflect + stretch, left 3, up 5}}

> **Q9: A rectangle has perimeter 40 m. Find the dimensions that maximize its area.** > > **Strategy Hint:** Area = $l \\times w$, with $2l + 2w = 40$. > > $2l + 2w = 40 \\implies l + w = 20 \\implies w = 20 - l$ > Area $A(l) = l(20 - l) = -l^2 + 20l$ > Vertex: $l_v = -\\frac{20}{2(-1)} = 10$, $w = 10$ > Maximum area: $10 \\times 10 = 100$ m² > >

\boxed{10\text{m} \times 10\text{m} \text{ (a square)}}

> **Q10: Determine how many $x$-intercepts $f(x) = 2x^2 - 3x + 5$ has.** > > **Strategy Hint:** Compute $D = b^2 - 4ac$. > > $D = (-3)^2 - 4(2)(5) = 9 - 40 = -31 < 0$ \to no real $x$-intercepts. > >

\boxed{0 \text{ (no real intercepts)}}

> **Q11: Give the standard form of $f(x) = 3(x - 2)^2 + 4$.** > > **Strategy Hint:** Expand: $a(x - h)^2 + k = a(x^2 - 2hx + h^2) + k$. > > $3(x - 2)^2 + 4 = 3(x^2 - 4x + 4) + 4 = 3x^2 - 12x + 12 + 4 = 3x^2 - 12x + 16$ > >

\boxed{f(x) = 3x^2 - 12x + 16}

> **Q12: The graph of $f(x) = x^2 + bx + c$ has vertex $(3, -1)$. Find $b$ and $c$.** > > **Strategy Hint:** Use $x_v = -b/(2a)$ and $f(x_v) = y_v$. > > $x_v = -\\frac{b}{2} = 3 \\implies b = -6$ > $f(3) = 9 + 3b + c = -1 \\implies 9 - 18 + c = -1 \\implies c = 8$ > >

\boxed{b = -6,\ c = 8}$$ --- ## 🔗 Cross-References - Next topic: [Quadratic Equations & Discriminant](quadratics-equations.md) - Related: [Polynomials](../week04/polynomial-operations.md) — quadratics are degree-2 polynomials - Across courses: BSMA1002 Stats 1 (quadratic regression); BSMA1003 Maths 2 (conic sections); gradient descent in ML optimizes quadratic-like loss functions Join Discord Previous2.2 Straight LinesNext3.2 Quadratic Equations & Discriminant