Welcome to Algebra and Functions! (Pure Mathematics 2)
Hello future mathematician! This chapter is incredibly important because it takes the fundamental algebra you learned in P1 and gives it some serious superpowers. We are going to dive deep into polynomials, learn brilliant shortcuts for finding remainders and factors, and explore how functions can be combined and reversed.
Don't worry if some of these ideas seem complex at first. We will break them down into simple, manageable steps. By the end of these notes, you'll be able to solve cubic equations and manipulate graphs like a pro!
Section 1: The Remainder and Factor Theorems
In P2, we focus heavily on expressions called polynomials, which are sums of terms involving powers of a variable (like \(x^3 - 4x + 1\)). Often, we need to divide one polynomial by another to find its factors or the remainder.
1.1 Understanding Polynomial Division
When you divide a polynomial \(P(x)\) by a divisor \(D(x)\), you get a quotient \(Q(x)\) and a remainder \(R\).
The relationship looks like this:
$$P(x) = D(x)Q(x) + R$$
Why do we need algebraic division?
If the remainder \(R\) is zero, it means the divisor \(D(x)\) is a factor of \(P(x)\). Finding factors is crucial for solving polynomial equations (like cubic equations).
(Note: While algebraic long division is a core skill, the Remainder and Factor Theorems below offer a much faster way to check for factors or find the remainder.)
1.2 The Remainder Theorem: The Shortcut
The Remainder Theorem is a fantastic shortcut! It tells you the remainder without doing the long division.
The Rule:
When a polynomial \(P(x)\) is divided by a linear divisor \((x-a)\), the remainder is always equal to \(P(a)\).
Step-by-Step Example:
- Identify the divisor, e.g., \((x-3)\). This means \(a=3\).
- Substitute this value into the polynomial \(P(x)\).
- The resulting number, \(P(3)\), is the remainder.
Common Mistake Alert!
If the divisor is \((x+5)\), then \(a = -5\). You must substitute the value that makes the bracket equal to zero.
1.3 The Factor Theorem: Finding Roots
The Factor Theorem is simply a special case of the Remainder Theorem.
The Rule:
If a polynomial \(P(x)\) is divided by \((x-a)\), and the remainder \(P(a)\) is zero, then \((x-a)\) is a factor of \(P(x)\).
Memory Aid: If \(P(a) = 0\), the remainder is zero, meaning \((x-a)\) fits perfectly—it's a factor!
Using the Factor Theorem to Solve Equations
This is the main application! If you have a cubic equation, you need to find at least one factor using trial and error (testing small, easy numbers like \(-2, -1, 1, 2\)).
Process for Solving Cubic Equations:
- Find the First Factor: Test simple values of \(a\) until you find one where \(P(a)=0\). Let's say you found \(P(2)=0\). This means \((x-2)\) is a factor.
- Find the Remaining Factor (The Quadratic): Divide \(P(x)\) by \((x-2)\) using algebraic long division or inspection. This will give you a quadratic expression, \(Q(x)\).
- Factorize Completely: Factorize the resulting quadratic \(Q(x)\) into two linear factors (if possible).
- Solve: Set all factors equal to zero to find the roots.
Key Takeaway for Section 1: The Remainder and Factor Theorems let us analyze polynomials quickly. The Remainder Theorem finds the leftover amount; the Factor Theorem confirms if \((x-a)\) fits perfectly (remainder = 0).
Section 2: Functions: Composition and Inverse
Functions are essentially mathematical processes. We put an input in, and we get a single, specific output. In P2, we learn how to chain these processes together (composition) and how to run them backwards (inverse).
2.1 Composite Functions (\(fg(x)\))
A composite function is formed when the output of one function becomes the input for another.
Notation:
$$fg(x)$$
This means apply \(g\) first, then apply \(f\) to the result.
Analogy: Imagine two machines. \(g\) is the first machine you feed the material to. Its output drops straight into machine \(f\).
Step-by-Step Calculation:
Let \(f(x) = 2x + 1\) and \(g(x) = x^2\).
To find \(fg(x)\):
- Start with the outer function: \(f(x) = 2x + 1\).
- Replace every \(x\) in \(f\) with the entire function \(g(x)\).
- $$fg(x) = 2(g(x)) + 1$$
- Substitute the definition of \(g(x)\): $$fg(x) = 2(x^2) + 1$$
Crucial Point: \(fg(x)\) is usually NOT the same as \(gf(x)\)!
2.2 Inverse Functions (\(f^{-1}(x)\))
The inverse function, \(f^{-1}(x)\), reverses the action of \(f(x)\). If \(f(2) = 5\), then \(f^{-1}(5) = 2\).
Analogy: If \(f(x)\) is "Add 3, then multiply by 2," the inverse \(f^{-1}(x)\) must be "Divide by 2, then subtract 3." Notice the order is reversed!
Finding the Inverse Function Algebraically
To find \(f^{-1}(x)\), follow these simple steps:
Example: Find the inverse of \(f(x) = 4x - 7\).
- Replace \(f(x)\) with \(y\):
$$y = 4x - 7$$ - Swap \(x\) and \(y\): (This mathematically reverses the roles of input and output)
$$x = 4y - 7$$ - Rearrange to make \(y\) the subject:
$$x + 7 = 4y$$
$$y = \frac{x + 7}{4}$$ - Write in inverse notation:
$$f^{-1}(x) = \frac{x + 7}{4}$$
Graphing Inverses:
The graph of \(y = f^{-1}(x)\) is always a reflection of the graph of \(y = f(x)\) across the line \(y = x\).
Did you know? A function must be a one-to-one mapping (meaning every output comes from only one input) in order to have a valid inverse over its whole domain. If it’s not, we have to restrict the domain (a common requirement for quadratic functions).
Key Takeaway for Section 2: Composition means applying functions sequentially (\(fg(x)\) is \(g\) then \(f\)). Inverse functions reverse the process, found by swapping \(x\) and \(y\) and rearranging.
Section 3: Transformations of Graphs
You studied basic transformations in P1. P2 requires you to master all four types—translation, stretching, and reflection—and understand the critical difference between affecting the \(x\) variable (inside the function) and the \(y\) variable (outside the function).
3.1 Understanding the General Rules
Let \(y = f(x)\) be the original curve.
The Golden Rule: Changes applied outside the bracket \((f(x) \pm a)\) affect the \(y\)-axis (vertical) and are intuitive. Changes applied inside the bracket \((f(x \pm a))\) affect the \(x\)-axis (horizontal) and are counter-intuitive (the opposite of what you might expect).
| Transformation | Rule | Effect |
|---|---|---|
| Vertical Translation (Up/Down) | \(y = f(x) + a\) | Shift graph up by \(a\) units (y-coordinate $\to$ y+a) |
| Horizontal Translation (Left/Right) | \(y = f(x + a)\) | Shift graph left by \(a\) units (x-coordinate $\to$ x-a). (Counter-intuitive!) |
| Vertical Stretch | \(y = a f(x)\) | Stretch graph vertically by scale factor \(a\) (y-coordinate $\to$ ay) |
| Horizontal Stretch | \(y = f(ax)\) | Stretch graph horizontally by scale factor \(1/a\). Shrinks if $a>1$. (x-coordinate $\to$ x/a). (Counter-intuitive!) |
| Reflection in x-axis | \(y = -f(x)\) | Reflects vertically. (y-coordinate $\to$ -y) |
| Reflection in y-axis | \(y = f(-x)\) | Reflects horizontally. (x-coordinate $\to$ -x) |
Encouragement: Don't worry if the horizontal shifts feel confusing! Just remember: if it's inside, it's a lie! \(+a\) inside means minus \(a\) movement on the x-axis.
3.2 Combining Transformations
You may be asked to describe the sequence of transformations required to get from \(y=f(x)\) to a complex function like \(y = 2f(x-3) + 1\).
The Order Matters! (Think SR-T)
When combining transformations, especially involving stretches/reflections and translations, the order is crucial.
Standard Order Rule:
Apply Stretches and Reflections (SR) first, followed by Translations (T).
Example: Describe the transformations from \(y = f(x)\) to \(y = 2f(x) + 5\).
- Stretch/Reflection (SR): Vertical stretch by scale factor 2 (multiplying the \(y\) values). (This comes from the 2 outside the function).
- Translation (T): Vertical translation of 5 units up (adding 5 to the \(y\) values). (This comes from the +5 outside the function).
A Note on Horizontal Order
If the transformation involves a coefficient on \(x\) (a horizontal stretch) combined with a horizontal translation, you sometimes need to factorize the function first to correctly identify the shifts.
For example, \(f(2x + 6)\) should be rewritten as \(f(2(x + 3))\) to correctly identify the stretch (sf 1/2) and the translation (3 units left).
Key Takeaway for Section 3: Know the difference between internal (\(x\)-axis, counter-intuitive) and external (\(y\)-axis, intuitive) changes. When combining steps, always apply Stretches/Reflections before Translations (SRT).
Chapter Review: Quick Check
Algebra and Functions Core P2 Skills:
- Remainder Theorem: Find remainder \(R\) when dividing by \((x-a)\) by calculating \(P(a)\).
- Factor Theorem: \((x-a)\) is a factor if \(P(a) = 0\).
- Composite Functions: \(fg(x)\) means substitute \(g(x)\) into \(f\).
- Inverse Functions: Found by setting \(y=f(x)\), swapping \(x\) and \(y\), and rearranging.
- Transformations: \(y = f(x-a)\) moves right. \(y = f(ax)\) stretches horizontally by factor \(1/a\). Order matters (SRT).