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Quadratic Functions

1. Core Concepts & Intuition

(If a straight line is the simplest path, a quadratic function describes the next simplest, and much more interesting, path: a curve. Think of the arc a ball makes when you throw it. It goes up, reaches a peak, and comes back down. That curve is a parabola, and it’s the shape of every quadratic function. Quadratics are used to model everything from projectile motion to optimizing profits, as they are the simplest functions that have a maximum or a minimum point.)

2. Formal Definitions, Jargon, and Nuances

A Quadratic Function is a polynomial function of degree 2, with the general form $f(x)=a{x}^2+bx+c$, where a, b, and c are constants and $a\ne 0$.

The graph of a quadratic function is a parabola. If $a>0$, the parabola opens upwards and has a minimum point. If $a<0$, it opens downwards and has a maximum point.

Key Features of a Parabola:

1. Vertex: The minimum or maximum point of the parabola. Its x-coordinate is given by $x=-\frac{b}{2a}$.
2. Axis of Symmetry: A vertical line that passes through the vertex, given by the equation $x=-\frac{b}{2a}$, dividing the parabola into two mirror images.
3. Roots (or x-intercepts): The points where the parabola crosses the x-axis. They are the solutions to the equation $a{x}^2+bx+c=0$ and can be found using the Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.
4. Discriminant ($D$ or $\Delta$): The part of the quadratic formula under the square root: $D=b^2-4ac$. It tells us the nature of the roots:
   • If $D>0$: Two distinct real roots.
   • If $D=0$: Exactly one real root (the vertex touches the x-axis).
   • If $D<0$: No real roots (the parabola never crosses the x-axis).

3. Step-by-Step Procedures & Worked Examples

Procedure: Analyzing and Graphing a Quadratic Function

1. Identify the coefficients a, b, and c from the equation $f(x)=a{x}^2+bx+c$.
2. Determine the direction of the parabola: upwards ($a>0$) or downwards ($a<0$).
3. Find the x-coordinate of the vertex using $x=-\frac{b}{2a}$.
4. Find the y-coordinate of the vertex by plugging the x-coordinate back into the function: $y=f(-\frac{b}{2a})$.
5. Find the roots (x-intercepts) by solving $a{x}^2+bx+c=0$ using the quadratic formula.
6. Find the y-intercept by setting $x=0$, which is always the point (0, c).
7. Plot the vertex, roots, and y-intercept, and draw a smooth parabola connecting them.

Example 1: Analyze and sketch $f(x)=x^2-4x+3$.

1. $a=1,b=-4,c=3$.
2. $a>0$, so it opens upwards.
3. Vertex x-coordinate: $x=-\frac{-4}{2(1)}=2$.
4. Vertex y-coordinate: $f(2)=(2{)}^2-4(2)+3=4-8+3=-1$. Vertex is at (2, -1).
5. Roots: $x=\frac{4\pm \sqrt{(-4{)}^2-4(1)(3)}}{2(1)}=\frac{4\pm \sqrt{16-12}}{2}=\frac{4\pm \sqrt{4}}{2}=\frac{4\pm 2}{2}$. The roots are $x=3$ and $x=1$.
6. y-intercept is (0, 3).

Visual Representation

(A precise plot of a parabola is best done with a graphing tool. The description would be: A parabola opening upwards, with its minimum point (vertex) at (2, -1). It crosses the x-axis at x=1 and x=3, and crosses the y-axis at y=3.)

4. The “Exam Brain” Algorithm & Strategic Handbook

Pattern Recognition:

• Keywords to look for: “parabola”, “vertex”, “maximum/minimum value”, “roots”, “axis of symmetry”, any equation with an $x^2$ term.
• Question Formats: “Find the vertex of the parabola…”, “For what value of x is the function maximized/minimized?”, “Find the roots of the quadratic equation…”, “How many real roots does the equation have?“.

Mental Algorithm (The Approach):

1. Identify the Goal: Am I looking for a specific point (vertex, roots), a value (max/min value), or a property (number of roots)?
2. Select the Tool:
   • Vertex, Max/Min, Axis of Symmetry? The vertex formula $x=-\frac{b}{2a}$ is your primary tool.
   • Roots? The quadratic formula is the most reliable tool. Factoring is faster but not always possible.
   • Number of roots? Use the discriminant, $D=b^2-4ac$. You don’t need to calculate the full quadratic formula.
3. Execute & Verify: Be careful with signs, especially with the -b in the formulas. If you find the roots, check that their midpoint is the x-coordinate of the vertex. It should be: $\frac{r_1+r_2}{2}=-\frac{b}{2a}$.

5. Common Pitfalls & Exam Traps

• Trap 1: Confusing Maximum/Minimum Value with Location. The minimum value of the function is the y-coordinate of the vertex, not the x-coordinate. The question “what is the minimum value?” is different from “at what x does the minimum occur?“.
• Trap 2: Sign Errors in the Quadratic Formula. The formula starts with $-b$. If b is already negative (e.g., -4), then $-b$ becomes positive (4). This is a very common source of error.

6. Practice Exercises (Scaffolded Difficulty)

Exercise 1 (Concept Check)

Does the function $f(x)=5-3x^2$ have a maximum or a minimum value? Do not calculate it.

Exercise 2 (Application)

Find the vertex and the roots of the quadratic function $f(x)=2x^2+5x-3$.

Exercise 3 (Qualifier-Style Synthesis)

A ball is thrown upwards, and its height in meters after t seconds is given by the function $h(t)=-5t^2+20t+1$. What is the maximum height the ball reaches, and how long does it take to reach that height?

7. Comprehensive Solutions to Exercises

Solution 1:

The function is $f(x)=-3x^2+5$. The coefficient of the $x^2$ term is $a=-3$. Since $a<0$, the parabola opens downwards, which means it has a maximum value.

Solution 2:

1. Identify the Goal: Find vertex and roots for $f(x)=2x^2+5x-3$. ($a=2,b=5,c=-3$)
2. Select the Tool: Vertex formula and quadratic formula.
3. Execute & Verify:
   • Vertex:
     • x-coordinate: $x=-\frac{b}{2a}=-\frac{5}{2(2)}=-\frac{5}{4}$.
     • y-coordinate: $f(-\frac{5}{4})=2(-\frac{5}{4}{)}^2+5(-\frac{5}{4})-3=2(\frac{25}{16})-\frac{25}{4}-3=\frac{25}{8}-\frac{50}{8}-\frac{24}{8}=-\frac{49}{8}$.
     • Vertex is at $(-\frac{5}{4},-\frac{49}{8})$.
   • Roots:
     • $x=\frac{-5\pm \sqrt{5^2-4(2)(-3)}}{2(2)}=\frac{-5\pm \sqrt{25+24}}{4}=\frac{-5\pm \sqrt{49}}{4}=\frac{-5\pm 7}{4}$.
     • The roots are $x_1=\frac{-5+7}{4}=\frac{2}{4}=\frac{1}{2}$ and $x_2=\frac{-5-7}{4}=-\frac{12}{4}=-3$.

Solution 3:

1. Identify the Goal: Find the maximum height (y-value of vertex) and the time it takes (x-value of vertex) for $h(t)=-5t^2+20t+1$.
2. Select the Tool: The question asks for a maximum, which points directly to finding the vertex.
3. Execute & Verify:
   • Here, the variable is t instead of x. $a=-5,b=20,c=1$.
   • Time to reach max height (t-coordinate of vertex):
     • $t=-\frac{b}{2a}=-\frac{20}{2(-5)}=-\frac{20}{-10}=2$ seconds.
   • Maximum height (h-coordinate of vertex):
     • $h(2)=-5(2{)}^2+20(2)+1=-5(4)+40+1=-20+40+1=21$ meters.
   • Answer: It takes 2 seconds to reach a maximum height of 21 meters.

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8. Connections & Further Learning

• [[Coordinate Geometry and Straight Lines]]
• [[Polynomials]]
• [[Calculus - Derivatives and Extrema]]

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