Maths Week 3: Quadratic Functions

This week, we explore Quadratic Functions. These are second-degree polynomial functions whose graphs form a characteristic curve called a parabola. Understanding quadratics is crucial for modeling various real-world phenomena, from the trajectory of a projectile to optimizing business revenue.

📚 Table of Contents

1. Fundamental Concepts
2. Question Pattern Analysis
3. Detailed Solutions by Pattern
4. Practice Exercises
5. Visual Learning: Mermaid Diagrams
6. Common Pitfalls & Traps
7. Quick Refresher Handbook

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1. Fundamental Concepts

🎯 The Quadratic Function

A quadratic function is defined by the formula:

$f(x)=a{x}^2+bx+c$

where $a,b,c$ are real numbers and, crucially, $a\ne 0$.

• Graph: The graph is a U-shaped curve called a parabola.
• Concavity (Direction):
  • If $a>0$, the parabola opens upwards. It has a minimum point.
  • If $a<0$, the parabola opens downwards. It has a maximum point.

📈 Key Characteristics of a Parabola

• Vertex: The turning point of the parabola. It is either the lowest point (minimum) or the highest point (maximum).

  • Vertex Coordinates $(h,k)$: The vertex is located at $x=h=-\frac{b}{2a}$. The y-coordinate is $k=f(h)=f(-\frac{b}{2a})$.
  • The formula for the vertex is: $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$

• Axis of Symmetry: A vertical line that passes through the vertex, dividing the parabola into two mirror images.

  • Equation: $x=-\frac{b}{2a}$

• Roots (or x-intercepts): These are the points where the parabola intersects the x-axis, i.e., where $f(x)=0$. They can be found using the Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

  • The term inside the square root, $\Delta =b^2-4ac$, is called the discriminant.
    • If $\Delta >0$, there are two distinct real roots.
    • If $\Delta =0$, there is exactly one real root (the vertex touches the x-axis).
    • If $\Delta <0$, there are no real roots (the parabola does not intersect the x-axis).

📐 The Slope of a Parabola

The slope of a parabola is not constant; it changes at every point. The slope at any given point $x$ on the curve $y=a{x}^2+bx+c$ is found using its derivative.

• Slope Formula: $\text{Slope}=f^{\prime }(x)=2ax+b$
• Slope at the Vertex: At the vertex ($x=-b/2a$), the slope is always zero, because the tangent line is horizontal. $2a\left(-\frac{b}{2a}\right)+b=-b+b=0$

📝 Vertex Form

Besides the standard form, the vertex form is another useful way to write a quadratic function: $f(x)=a(x-h{)}^2+k$ where $(h,k)$ are the coordinates of the vertex. This form is excellent for problems where the vertex or axis of symmetry is known.

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2. Question Pattern Analysis

From the provided assignment files, we can identify four primary patterns for questions related to quadratic functions.

🎯 Pattern 1: Finding Coefficients from Slopes

• Frequency: High
• Description: Problems provide the slope of a parabola at one or more points and ask you to determine the coefficients ($a$, $b$, or $c$).
• Core Skill: Using the slope formula $f^{\prime }(x)=2ax+b$ to create a system of linear equations.
• Source: Maths/Week_3_Graded_Assignment_3_Sep_2025_Mathematics_I_Qualifier.md (Questions 1, 10).

🎯 Pattern 2: Solving Real-World Problems

• Frequency: High
• Description: Word problems involving quantities that have a quadratic relationship (e.g., area, projectile motion, number arrangements).
• Core Skill: Translating the problem statement into a quadratic equation and solving it.
• Source: Maths/Week_3_Graded_Assignment_3_Sep_2025_Mathematics_I_Qualifier.md (Questions 2, 3, 6).

🎯 Pattern 3: Maxima and Minima

• Frequency: High
• Description: Questions asking for the maximum or minimum value of a function, or the point at which it occurs.
• Core Skill: Finding and interpreting the vertex of the parabola.
• Source: Maths/Week_3_Graded_Assignment_3_Sep_2025_Mathematics_I_Qualifier.md (Questions 7, 8, 11).

🎯 Pattern 4: Intersections of Curves

• Frequency: Medium
• Description: Finding the point(s) where a parabola intersects with another parabola or a straight line.
• Core Skill: Setting the two functions’ equations equal to each other and solving the resulting equation.
• Source: Maths/Week_3_Graded_Assignment_3_Sep_2025_Mathematics_I_Qualifier.md (Questions 5, 9, 13).

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3. Detailed Solutions by Pattern

🎯 Pattern 1: Finding Coefficients from Slopes

Example (from Graded Assignment Q1):

If the slope of parabola $y=a{x}^2+bx+c$ at points $(3,2)$ and $(2,3)$ are $32$ and $2$ respectively, then find the value of $a$.

Step-by-Step Solution:

1. Write the Slope Formula: The slope of the parabola at any point $x$ is given by its derivative: $\text{Slope}=2ax+b$

2. Create Equations from Given Conditions:

   • At the point $(3,2)$, we have $x=3$ and the slope is 32. $2a(3)+b=32⟹6a+b=32\text{(Eq. 1)}$
   • At the point $(2,3)$, we have $x=2$ and the slope is 2. $2a(2)+b=2⟹4a+b=2\text{(Eq. 2)}$ (Note: The y-coordinates (2 and 3) and the coefficient c are not needed for this part of the problem.)

3. Solve the System of Linear Equations: We can solve for $a$ by subtracting the second equation from the first. $(6a+b)-(4a+b)=32-2$ $2a=30$ $a=15$

Answer: 15

🎯 Pattern 2: Solving Real-World Problems

Example (from Graded Assignment Q2):

A class of $140$ students are arranged in rows such that the number of students in a row is one less than thrice the number of rows. Find the number of students in each row.

Step-by-Step Solution:

1. Define Variables:

   • Let $r$ be the number of rows.
   • Let $s$ be the number of students in each row.

2. Translate the Word Problem into Equations:

   • “The number of students in a row ($s$) is one less than thrice the number of rows ($r$)”: $s=3r-1\text{(Eq. 1)}$
   • The total number of students is the number of rows times the students per row: $r\times s=140\text{(Eq. 2)}$

3. Create and Solve the Quadratic Equation:

   • Substitute the expression for $s$ from Eq. 1 into Eq. 2: $r(3r-1)=140$
   • Expand and rearrange into standard quadratic form ($a{x}^2+bx+c=0$): $3r^2-r=140$ $3r^2-r-140=0$
   • Solve this quadratic equation using the quadratic formula $r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$: $r=\frac{-(-1)\pm \sqrt{(-1{)}^2-4(3)(-140)}}{2(3)}$ $r=\frac{1\pm \sqrt{1+1680}}{6}$ $r=\frac{1\pm \sqrt{1681}}{6}$ $r=\frac{1\pm 41}{6}$
   • This gives two possible solutions: $r=\frac{42}{6}=7$ or $r=\frac{-40}{6}$ (which is negative).

4. Interpret the Result: Since the number of rows cannot be negative, we must have $r=7$.

5. Find the Number of Students per Row:

   • Using Eq. 1: $s=3r-1=3(7)-1=21-1=20$.

Answer: 20

🎯 Pattern 3: Maxima and Minima

Example (from Graded Assignment Q8):

A water fountain’s stream is modeled by $h(t)=-0.5t^2+4t+1$, where $h(t)$ is height in meters and $t$ is time in seconds. Determine the time (in seconds) it takes for the water to reach its maximum height.

Step-by-Step Solution:

1. Identify the Goal: “Maximum height” corresponds to the vertex of the parabola. The “time it takes” is the x-coordinate (or in this case, the t-coordinate) of the vertex.

2. Identify Coefficients: From $h(t)=-0.5t^2+4t+1$, we have:

   • $a=-0.5$
   • $b=4$
   • $c=1$ (Since $a<0$, the parabola opens downwards, so it has a maximum.)

3. Apply the Vertex Formula for the t-coordinate: $t=-\frac{b}{2a}$ $t=-\frac{4}{2(-0.5)}$ $t=-\frac{4}{-1}$ $t=4$

Answer: 4 seconds

🎯 Pattern 4: Intersections of Curves

Example (from Graded Assignment Q9):

Find the intersection points of the curve $y=4x^2+x+6$ and the straight line passing through the points $(1,6)$ and $(4,5)$.

Step-by-Step Solution:

1. Find the Equation of the Straight Line:

   • First, find the slope ($m$): $m=\frac{y_2-y_1}{x_2-x_1}=\frac{5-6}{4-1}=-\frac{1}{3}$
   • Use the point-slope form $y-y_1=m(x-x_1)$ with point $(1,6)$: $y-6=-\frac{1}{3}(x-1)$ $y=-\frac{1}{3}x+\frac{1}{3}+6$ $y=-\frac{1}{3}x+\frac{19}{3}$

2. Set the Equations Equal to Each Other: To find the intersection points, we set the y-values equal: $4x^2+x+6=-\frac{1}{3}x+\frac{19}{3}$

3. Solve the Resulting Quadratic Equation:

   • Multiply everything by 3 to eliminate fractions: $12x^2+3x+18=-x+19$
   • Rearrange into standard form ($a{x}^2+bx+c=0$): $12x^2+4x-1=0$
   • Use the quadratic formula to solve for $x$: $x=\frac{-4\pm \sqrt{4^2-4(12)(-1)}}{2(12)}$ $x=\frac{-4\pm \sqrt{16+48}}{24}$ $x=\frac{-4\pm \sqrt{64}}{24}=\frac{-4\pm 8}{24}$
   • This gives two solutions for $x$:
     • $x_1=\frac{-4+8}{24}=\frac{4}{24}=\frac{1}{6}$
     • $x_2=\frac{-4-8}{24}=\frac{-12}{24}=-\frac{1}{2}$

4. Find the Corresponding y-coordinates: Substitute these x-values back into the line equation (it’s simpler).

   • For $x_1=1/6$: $y_1=-\frac{1}{3}\left(\frac{1}{6}\right)+\frac{19}{3}=-\frac{1}{18}+\frac{114}{18}=\frac{113}{18}$
   • For $x_2=-1/2$: $y_2=-\frac{1}{3}\left(-\frac{1}{2}\right)+\frac{19}{3}=\frac{1}{6}+\frac{38}{6}=\frac{39}{6}=\frac{13}{2}$

Answer: The intersection points are $(\frac{1}{6},\frac{113}{18})$ and $(-\frac{1}{2},\frac{13}{2})$.

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4. Practice Exercises

Exercise 1: Maximum Value and Vertex

A company’s daily profit, $P(x)$, from selling $x$ units of a product is given by $P(x)=-2x^2+120x-800$. a) How many units should be sold to maximize profit? b) What is the maximum possible daily profit?

Hint

The maximum profit occurs at the vertex of the parabola. Part (a) asks for the x-coordinate and part (b) asks for the y-coordinate of the vertex.

Solution

Here, $a=-2$, $b=120$, $c=-800$. a) The number of units to maximize profit is the x-coordinate of the vertex: $$ x = -\frac{b}{2a} = -\frac{120}{2(-2)} = -\frac{120}{-4} = 30 \text{ units} $$ b) The maximum profit is the y-coordinate of the vertex, $P(30)$: $$ P(30) = -2(30)^2 + 120(30) - 800 $$ $$ P(30) = -2(900) + 3600 - 800 $$ $$ P(30) = -1800 + 3600 - 800 = 1000 $$ **Answers:** a) 30 units, b) ₹1000.

Exercise 2: Finding Roots

The product of two consecutive odd natural numbers is 143. Find the numbers.

Hint

Let the first odd number be $n$. The next consecutive odd number is $n+2$. Their product is $n(n+2) = 143$. Solve this quadratic equation.

Solution

1. **Set up the equation:** Let the numbers be $n$ and $n+2$. $$ n(n+2) = 143 $$ $$ n^2 + 2n - 143 = 0 $$ 2. **Solve the quadratic equation:** Using the quadratic formula: $$ n = \frac{-2 \pm \sqrt{2^2 - 4(1)(-143)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 572}}{2} = \frac{-2 \pm \sqrt{576}}{2} = \frac{-2 \pm 24}{2} $$ 3. **Choose the valid solution:** This gives $n = \frac{22}{2} = 11$ or $n = \frac{-26}{2} = -13$. Since the question asks for "natural numbers", we choose the positive solution, $n=11$. 4. **Find the second number:** The second number is $n+2 = 11+2 = 13$. **Answer:** The numbers are 11 and 13.

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5. Visual Learning: Mermaid Diagrams

Anatomy of a Parabola

graph TD
    A[y = ax^2+bx+c] --> B{a > 0?};
    B -- Yes --> C(Opens Upward);
    B -- No --> D(Opens Downward);
    C --> E[Vertex is a Minimum];
    D --> F[Vertex is a Maximum];
    A --> G[Axis of Symmetry: x = -b/2a];
    A --> H[Roots: x = (-b ± sqrt(b^2-4ac))/2a];
    G --> I[Vertex: (-b/2a, f(-b/2a))];

Problem-Solving Flow for Intersections

graph LR
    A[Start: Two functions y=f(x), y=g(x)] --> B{Set f(x) = g(x)};
    B --> C[Rearrange into a single equation];
    C --> D[Solve for x];
    D --> E{Found x values?};
    E -- Yes --> F[Substitute x back into f(x) or g(x) to find y];
    F --> G[End: Intersection points (x,y)];
    E -- No --> H[End: No intersection];

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6. Common Pitfalls & Traps

• Forgetting $a\ne 0$: The function is only quadratic if the $x^2$ term exists. If $a=0$, it becomes a linear function.
• Sign Errors: The formula for the vertex x-coordinate is $x=-b/(2a)$. A very common mistake is to forget the negative sign, or to handle a negative $b$ value incorrectly (e.g., $-(-5)$ becomes $5$).
• Maximum vs. Minimum Confusion: Remember this simple rule:
  • a > 0 $\to$ Positive $\to$ Happy face U $\to$ Minimum value.
  • a < 0 $\to$ Negative $\to$ Sad face ∩ $\to$ Maximum value.
• Incomplete Answers: When asked for intersection points, providing only the x-values is an incomplete answer. You must find the corresponding y-values to give the full coordinates.

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7. Quick Refresher Handbook

Concept	Formula / Definition	Purpose & Key Insight
Standard Form	$f(x)=a{x}^2+bx+c$	The basic form. The sign of ‘a’ tells you the direction.
Vertex Form	$f(x)=a(x-h{)}^2+k$	Immediately tells you the vertex is at $(h,k)$.
Vertex x-coordinate	$h=-\frac{b}{2a}$	Finds the axis of symmetry and the location of the max/min.
Max/Min Value	$k=f(-\frac{b}{2a})$	The highest or lowest value the function can achieve.
Quadratic Formula	$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$	The ultimate tool for finding the roots (x-intercepts).
Discriminant	$\Delta =b^2-4ac$	Quickly tells you how many real roots exist (2, 1, or 0).
Slope at a Point	$f^{\prime }(x)=2ax+b$	Calculates the steepness of the parabola at any x-value.