Maths1_Week3_Quadratic_Functions

Maths 1 Week 3: Quadratic Functions & Equations

0. Prerequisites

NOTE

What you need to know:

• Solving Linear Equations: Finding $x$ in $3x+5=0$.
• Basic Factoring: Knowing that $x^2-4=(x-2)(x+2)$.
• Graphing: Plotting points on X-Y plane.

Quick Refresher

• Polynomial: Expression like $a{x}^n+\dots$.
• Degree: Highest power of $x$. Quadratic has degree 2 ($x^2$).
• Roots: Values of $x$ where the equation equals zero.

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1. Core Concepts

1.1 Quadratic Functions

A function of the form $f(x)=a{x}^2+bx+c$ ($a\ne 0$).

• Graph: A Parabola (U-shape).
  • If $a>0$: Opens Up (Has a Minimum).
  • If $a<0$: Opens Down (Has a Maximum).

1.2 Key Features

• Vertex ($V$): The turning point (tip) of the parabola.
  • x-coordinate: $x_v=\frac{-b}{2a}$
  • y-coordinate: $y_v=f(x_v)=c-\frac{b^2}{4a}$
• Axis of Symmetry: Vertical line passing through vertex: $x=\frac{-b}{2a}$.
• Discriminant ($D$): $D=b^2-4ac$.
  • $D>0$: 2 Real Roots (Cuts X-axis twice).
  • $D=0$: 1 Real Root (Touches X-axis).
  • $D<0$: No Real Roots (Floats above/below X-axis).

1.3 Quadratic Formula

To solve $a{x}^2+bx+c=0$: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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2. Pattern Analysis & Goated Solutions

Pattern 1: Optimization (Max/Min Problems)

Context: “Find the maximum height”, “Find time to reach max height”, “Maximize area”.

TIP

Mental Algorithm:

1. Identify: Is it a quadratic ($t^2$ term)?
2. Check Sign: Is $a$ negative? (Expect Max). Is $a$ positive? (Expect Min).
3. Find Vertex X: Calculate $t=\frac{-b}{2a}$. This is when it happens.
4. Find Vertex Y: Plug $t$ back into equation. This is the value (Max Height).

Example (Detailed Solution)

Problem: A ball’s height is given by $h(t)=-5t^2+20t+2$. Find the maximum height and when it occurs. Solution:

1. Identify: Quadratic with $a=-5,b=20,c=2$.
2. Check: $a=-5$ (Negative), so it opens down. We have a Maximum.
3. Find Time ($t$):
   • $t=\frac{-b}{2a}=\frac{-20}{2(-5)}$.
   • $t=\frac{-20}{-10}=2$ seconds.
   • Meaning: The ball reaches peak at 2 seconds.
4. Find Height ($h$):
   • Plug $t=2$ into $h(t)$.
   • $h(2)=-5(2{)}^2+20(2)+2$.
   • $h(2)=-5(4)+40+2$.
   • $h(2)=-20+42=22$ meters. Answer: Max height is 22m at 2 seconds.

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Pattern 2: Intersection of Line and Parabola

Context: “Find where the line meets the curve”, “Solve the system”.

TIP

Mental Algorithm:

1. Equate: Set $y_{line}=y_{parabola}$.
2. Rearrange: Move everything to one side to get $A{x}^2+Bx+C=0$.
3. Solve: Find $x$ values using formula or factoring.
4. Plug Back: Find corresponding $y$ values using the linear equation (easier).

Example (Detailed Solution)

Problem: Find intersection of $y=x^2$ and $y=x+2$. Solution:

1. Equate: $x^2=x+2$.
2. Rearrange: $x^2-x-2=0$.
3. Solve:
   • Factor: Find numbers that multiply to -2 and add to -1. (-2, 1).
   • $(x-2)(x+1)=0$.
   • $x=2$ or $x=-1$.
4. Find Y:
   • If $x=2$: $y=2+2=4$. Point $(2,4)$.
   • If $x=-1$: $y=-1+2=1$. Point $(-1,1)$. Answer: Intersects at $(2,4)$ and $(-1,1)$.

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Pattern 3: Word Problems (Number Properties)

Context: “Product of two consecutive odd numbers is…”

TIP

Mental Algorithm:

1. Define Variables:
   • Consecutive integers: $x,x+1$.
   • Consecutive ODD/EVEN: $x,x+2$.
2. Equation: Write the product equation.
3. Solve: Solve the quadratic. Discard invalid answers (e.g., if asked for natural numbers, ignore negatives).

Example (Detailed Solution)

Problem: Product of two consecutive positive even integers is 48. Find them. Solution:

1. Define: Let numbers be $x$ and $x+2$.
2. Equation: $x(x+2)=48$.
3. Expand: $x^2+2x=48⟹x^2+2x-48=0$.
4. Solve:
   • Factors of -48 adding to 2: (8, -6).
   • $(x+8)(x-6)=0$.
   • $x=-8$ or $x=6$.
5. Filter: Question asks for “positive”. Discard -8.
   • So $x=6$.
   • Next number $x+2=8$. Answer: The numbers are 6 and 8.

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3. Practice Exercises

1. Vertex: Find vertex of $y=x^2-6x+5$.
   • Hint: $x=-(-6)/2=3$. Plug in 3.
2. Roots: Solve $2x^2-8=0$.
   • Hint: $2x^2=8⟹x^2=4$.
3. Intersection: Line $y=4$ intersects $y=x^2$. Where?
   • Hint: $x^2=4⟹x=\pm 2$. Points $(2,4),(-2,4)$.

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🧠 Level Up: Advanced Practice

Question 1: Common Roots Condition

Problem: Find $k$ if $x^2+kx+6=0$ and $x^2-5x+6=0$ have a common root. Logic:

1. Solve the known equation: $x^2-5x+6=0⟹(x-2)(x-3)=0$. Roots are 2, 3.
2. Case 1: Common root is 2.
   • Substitute $x=2$ into first eq: $4+2k+6=0⟹2k=-10⟹k=-5$.
3. Case 2: Common root is 3.
   • Substitute $x=3$ into first eq: $9+3k+6=0⟹3k=-15⟹k=-5$. Answer: $k=-5$. (In this case, both roots are actually common!).

Question 2: Optimization with Constraints

Problem: Maximize Area $A=x\cdot y$ subject to $2x+y=100$. Logic:

1. Express one variable: $y=100-2x$.
2. Substitute: $A(x)=x(100-2x)=100x-2x^2$.
3. Find Vertex: This is a downward parabola ($a=-2$).
   • Max occurs at $x=-b/2a=-100/(2\cdot -2)=25$.
4. Find y: $y=100-2(25)=50$.
5. Max Area: $25\times 50=1250$. Answer: 1250.

Question 3: Intersection of Parabolas

Problem: Find intersection points of $y=x^2$ and $y=2-x^2$. Logic:

1. Equate: $x^2=2-x^2$.
2. Solve: $2x^2=2⟹x^2=1⟹x=\pm 1$.
3. Find y: If $x=1,y=1$. If $x=-1,y=1$. Answer: $(1,1)$ and $(-1,1)$.