Chapter 2: Algebra and Graphs

Worked Example: Simplifying an Algebraic Fraction

Let's Cancel This Fraction Like a Bad Subscription ✂️

1. 1
   Okay, our mission is to simplify this fraction. The only way to do that is to see if the top (numerator) and bottom (denominator) share any common factors. So, our first move is to factorise both expressions.

   Goal: Factorise $x^2 + 2x - 8$ and $x^2 - 4$.
2. 2
   Let's start with the numerator, $x^2 + 2x - 8$. This is a standard quadratic. We need two numbers that multiply to make -8 and add to make +2. Let's think... +4 and -2 work! ($4 \times -2 = -8$ and $4 + (-2) = 2$).

   $x^2 + 2x - 8 = (x+4)(x-2)$
3. 3
   Now for the denominator, $x^2 - 4$. You should immediately spot this as a classic 'Difference of Two Squares' pattern ($a^2 - b^2$). Here, $a=x$ and $b=2$ (since $2^2=4$). This factorises into $(a-b)(a+b)$.

   $x^2 - 4 = x^2 - 2^2 = (x-2)(x+2)$
4. 4
   Now we can rewrite the whole fraction using our new factorised versions. Look for any identical brackets on the top and bottom that we can cancel out. It looks like we have an $(x-2)$ on both!

   $\frac{(x+4)(x-2)}{(x-2)(x+2)} \implies \frac{x+4}{x+2}$
5. 5
   After cancelling the common factor, we're left with our final, simplified answer. Way cleaner than what we started with! Job done.

   $\frac{x+4}{x+2}$

Worked Example: Solving Simultaneous Equations (Linear and Non-Linear)

When a Line & a Parabola Go on a Date... 💘

1. 1
   Both equations are equal to 'y', so we can set them equal to each other. This is the substitution method. We're basically mashing the two equations together to eliminate 'y' and focus on finding 'x'.

   $x^2 + 5x - 2 = x + 3$
2. 2
   Now, we need to rearrange this into a standard quadratic equation ($ax^2 + bx + c = 0$). Let's move everything from the right side over to the left by subtracting 'x' and '3' from both sides.

   $x^2 + 5x - x - 2 - 3 = 0 \\ x^2 + 4x - 5 = 0$
3. 3
   Time to solve this quadratic! Factorisation looks like the easiest method here. We need two numbers that multiply to give -5 and add to give +4. Those numbers are +5 and -1.

   $(x+5)(x-1) = 0 \\ \text{So, } x = -5 \text{ or } x = 1$
4. 4
   We're not done! We have the x-coordinates, but we need the matching y-coordinates for our intersection points. Let's substitute each x-value back into the easiest equation, which is the linear one ($y = x + 3$).

   $\text{When } x = -5: \quad y = (-5) + 3 = -2 \\ \text{When } x = 1: \quad y = (1) + 3 = 4$
5. 5
   Finally, write down your solutions as coordinate pairs. These are the two points where the line and the parabola intersect on a graph. Job done!

   $\text{The solutions are } (-5, -2) \text{ and } (1, 4).$

Worked Example: Graphing Multiple Inequalities

Defining Your Weekend Grind Zone 🎮📚

1. 1
   First, let's translate the rules into three inequalities. 'Maximum of 8 hours total' means $x+y$ is less than or equal to 8. 'At least 2 hours studying' means $y$ is greater than or equal to 2. 'More time gaming than studying' means $x$ is greater than $y$.

   $x + y \le 8$ $y \ge 2$ $x > y$
2. 2
   Next, we'll draw the boundary lines for each inequality. We'll treat them as equations for now: $x+y=8$, $y=2$, and $x=y$. The first two will be solid lines because they are inclusive ($\le, \ge$), but $x=y$ will be a dashed line because it's a strict inequality ($>$).
   
   [IMAGE_PLACEHOLDER_2: A Cartesian graph with the x-axis labelled 'Gaming Hours (x)' and y-axis labelled 'Studying Hours (y)'. The three lines y=2, x=y, and x+y=8 are drawn. y=2 is a horizontal solid line. x=y is a diagonal dashed line. x+y=8 is a diagonal solid line.]

   (See graph above)
3. 3
   Now for the shading! We shade the unwanted region for each line.
   👉 For $x+y \le 8$, we want the area below the line, so we shade everything above it.
   👉 For $y \ge 2$, we want the area above the line, so we shade everything below it.
   👉 For $x > y$, we want the area where x-coordinates are bigger (to the right/below the line $y=x$), so we shade the region above it.

   Test a point like (4,3) to check. $4+3 \le 8$ (True). $3 \ge 2$ (True). $4 > 3$ (True). So the point (4,3) should be in our final unshaded region.
4. 4
   The area left unshaded is the solution region. It's a small triangle where all three conditions are met. We'll label this region 'R'. Any coordinate pair inside this triangle is a valid combination of gaming and study hours for your weekend. Go you! ✅
   
   [IMAGE_PLACEHOLDER_3: The same graph as before, but now with the unwanted regions shaded. The region above x+y=8 is shaded. The region below y=2 is shaded. The region above and to the left of y=x is shaded. A clear, unshaded triangle is left in the middle, which is labelled 'R'.]

   (See graph above)

Worked Example: Finding the nth Term of a Quadratic Sequence

Let's Level Up: Boss Battle with a Quadratic Sequence 🎮

1. 1
   First, we find the first differences between the terms to see if it's a linear sequence. If the differences aren't the same, we know we're dealing with something more complex.

   $15-4=11, \quad 32-15=17, \quad 55-32=23, \quad 84-55=29$
2. 2
   Since the first differences (11, 17, 23, 29) are not constant, we find the second differences. A constant second difference is the signature move of a quadratic sequence.

   $17-11=6, \quad 23-17=6, \quad 29-23=6$
3. 3
   The second difference is 6. For any quadratic sequence $an^2 + bn + c$, the second difference is always equal to $2a$. We use this to find our 'a' value.

   $2a = 6 \implies a = 3$
4. 4
   Now we know the sequence involves $3n^2$. We'll subtract the values of $3n^2$ from our original terms ($T_n$) to see what's left over. This will reveal a simpler, linear sequence.

   $T_n - 3n^2: \\ n=1: 4 - 3(1^2) = 1 \\ n=2: 15 - 3(2^2) = 3 \\ n=3: 32 - 3(3^2) = 5$
5. 5
   The remaining sequence is 1, 3, 5... This is a simple linear sequence. It goes up by 2 each time, so its rule is $2n$. To get from $2n$ to our sequence (e.g., for n=1, $2(1)=2$, but we need 1), we subtract 1. So, the rule is $2n-1$.

   $\text{Linear part} = 2n-1$
6. 6
   Finally, we combine the quadratic part and the linear part to get our complete formula. Always a good idea to test it with a term to make sure you've nailed it!

   $T_n = 3n^2 + 2n - 1$

Worked Example: Inverse Square Proportion

Solving a Gaming Headset Problem 🎧

1. 1
   First, we need to translate the sentence into math. 'Volume is inversely proportional to the square of the distance' becomes an algebraic relationship. Then, we introduce our trusty constant of proportionality, $k$, to make it a full equation.

   $V \propto \frac{1}{d^2} \implies V = \frac{k}{d^2}$
2. 2
   Now we need to find the value of $k$. We do this by plugging in the pair of values we already know: $V = 90$ when $d = 2$. This will lock in the specific relationship for this headset.

   $90 = \frac{k}{2^2} \implies 90 = \frac{k}{4} \implies k = 90 \times 4 \implies k = 360$
3. 3
   We've found our constant! Now we can write the complete formula that connects volume and distance for this specific situation. This is our rule for solving the rest of the problem.

   $V = \frac{360}{d^2}$
4. 4
   Finally, we use our formula to answer the question. We want to find the volume ($V$) when the distance is 6 cm. Just substitute $d=6$ into our equation and solve.

   $V = \frac{360}{6^2} = \frac{360}{36} = 10$
5. 5
   State the final answer clearly with its units. The volume is 10 decibels. This makes sense – the distance tripled, so the volume dropped significantly because of the inverse square relationship.

   The volume is 10 decibels.

Worked Example: Analysing a Speed-Time Graph

Let's Break Down This Journey 🚗💨

1. 1
   First, let's find the acceleration. On a speed-time graph, acceleration is the gradient of the line. We need to calculate the 'rise over run' for the first section of the journey.

   $\text{Acceleration} = \frac{\text{Change in Speed}}{\text{Time Taken}} = \frac{30 \text{ m/s} - 0 \text{ m/s}}{20 \text{ s}} = 1.5 \text{ m/s}^2$
2. 2
   Next, we need the total distance. Remember, distance is the total area under the speed-time graph. The graph forms a trapezium. Before we calculate, let's make sure all our time units are the same. The middle section is 2 minutes, which is $2 \times 60 = 120$ seconds.

   $\text{Total Time} = 20\text{s} + 120\text{s} + 30\text{s} = 170\text{s}$
3. 3
   We can now find the area of the trapezium. The parallel sides are the top flat section (the time at constant speed, 120s) and the bottom section (the total time, 170s). The height is the maximum speed (30 m/s). The formula for the area of a trapezium is $\frac{1}{2}(a+b)h$.

   $\text{Distance} = \frac{1}{2} \times (120 + 170) \times 30$
4. 4
   Let's finish the calculation to get our final answer. Don't forget the units!

   $\text{Distance} = \frac{1}{2} \times (290) \times 30 = 145 \times 30 = 4350 \text{ meters}$

Worked Example: Finding and Classifying Stationary Points

Let's Find Those Turning Points! 🚀

1. 1
   First, we need the 'gradient formula' for our curve. We get this by differentiating the equation for $y$ with respect to $x$. We'll apply the $anx^{n-1}$ rule to each term.

   $\frac{dy}{dx} = 3x^{3-1} + (2 \times 3)x^{2-1} - 9x^{1-1} + 0$
   $\frac{dy}{dx} = 3x^2 + 6x - 9$
2. 2
   Stationary points occur where the gradient is zero. So, we set our derivative $\frac{dy}{dx}$ equal to 0 and solve the resulting quadratic equation for $x$.

   $3x^2 + 6x - 9 = 0$
   Divide by 3 to simplify: $x^2 + 2x - 3 = 0$
   Factorise the quadratic: $(x+3)(x-1) = 0$
   This gives us two solutions: $x = -3 \text{ and } x = 1$
3. 3
   Now we have the x-coordinates, but we need the full $(x, y)$ coordinates. We find the corresponding y-values by substituting our x-values back into the original equation for $y$.

   For $x = -3$:
   $y = (-3)^3 + 3(-3)^2 - 9(-3) + 1 = -27 + 27 + 27 + 1 = 28$
   So one point is $(-3, 28)$.
   
   For $x = 1$:
   $y = (1)^3 + 3(1)^2 - 9(1) + 1 = 1 + 3 - 9 + 1 = -4$
   So the other point is $(1, -4)$.
4. 4
   Time to classify them! Are they hills or valleys? We'll use the second derivative test. We differentiate our first derivative, $\frac{dy}{dx}$, to find the second derivative, $\frac{d^2y}{dx^2}$.

   $\frac{dy}{dx} = 3x^2 + 6x - 9$
   $\frac{d^2y}{dx^2} = 6x + 6$
5. 5
   Finally, we substitute our x-coordinates into the second derivative. If $\frac{d^2y}{dx^2} < 0$, it's a maximum. If $\frac{d^2y}{dx^2} > 0$, it's a minimum.

   At $x = -3$:
   $\frac{d^2y}{dx^2} = 6(-3) + 6 = -18 + 6 = -12$
   Since $-12 < 0$, the point $(-3, 28)$ is a maximum.
   
   At $x = 1$:
   $\frac{d^2y}{dx^2} = 6(1) + 6 = 12$
   Since $12 > 0$, the point $(1, -4)$ is a minimum.

Worked Example: Composite and Inverse Functions

Let's Get This Bread: A Function Mash-Up 🍞

1. 1
   For part (a), we just need to find $f(5)$. This is a simple substitution. We're plugging the number 5 into the 'x' slot of our function $f(x)$.

   $f(5) = 3(5) - 2 = 15 - 2 = 13$
2. 2
   For part (b), we're finding the composite function $gf(x)$, which means $g(f(x))$. We take the entire expression for $f(x)$, which is $(3x-2)$, and substitute it into the 'x' in the $g(x)$ function.

   $gf(x) = g(3x - 2) = (3x - 2)^2 + 4 = (9x^2 - 12x + 4) + 4 = 9x^2 - 12x + 8$
3. 3
   Finally, for part (c), we need to find the inverse of $f(x)$. We'll follow the three-step process: 1. Set to y, 2. Swap x and y, 3. Solve for the new y.

   $\text{Step 1: Let } y = 3x - 2$ $\text{Step 2: Swap } x \text{ and } y \rightarrow x = 3y - 2$ $\text{Step 3: Solve for y } \rightarrow x + 2 = 3y \rightarrow y = \frac{x+2}{3}$ $\text{So, } f^{-1}(x) = \frac{x+2}{3}$

Worked Example: Constructing and Substituting into a Formula

1. 1
   For (a), spot the fixed bit ($6$) and the per-hour bit ($2.50$ per hour, so $2.50h$). Stick them together.

   $C = 6 + 2.50h$
2. 2
   For (b), sub $h = 4$ in. Don't rush — BODMAS: multiply first, then add.

   $C = 6 + 2.50 \times 4 = 6 + 10 = \$16$
3. 3
   For (c), bracket the negative before squaring. This is where marks get lost.

   $5 - 3 \times (-2)^2 = 5 - 3 \times 4$
4. 4
   BODMAS again — multiplication before subtraction.

   $= 5 - 12 = -7$

Worked Example: Fractional Indices and an Index Equation

1. 1
   For (a), use $a^{m/n} = (\sqrt[n]{a})^m$. Take the root first (smaller numbers = fewer mistakes), then raise.

   $16^{\frac{3}{4}} = (\sqrt[4]{16})^3 = 2^3 = 8$
2. 2
   For (b), the negative flips the fraction upside down, the half means square root. Do the flip first.

   $\left(\frac{1}{9}\right)^{-\frac{1}{2}} = 9^{\frac{1}{2}} = \sqrt{9} = 3$
3. 3
   For (c), get the RHS to the same base as the LHS. Spot $125 = 5^3$.

   $5^{2x-1} = 5^3$
4. 4
   Same base means the exponents must be equal. Solve the mini linear equation.

   $2x - 1 = 3 \implies 2x = 4 \implies x = 2$

Worked Example: Simplifying an Algebraic Fraction with Quadratics

1. 1
   Top is a difference of two squares: $a^2 - b^2 = (a-b)(a+b)$. Here $a = x$, $b = 2$.

   $x^2 - 4 = (x-2)(x+2)$
2. 2
   Bottom is a quadratic. Find two numbers that multiply to $+6$ and add to $-5$. Those are $-2$ and $-3$.

   $x^2 - 5x + 6 = (x-2)(x-3)$
3. 3
   Now cancel the common factor $(x-2)$. Same energy as cancelling a $5$ from $\tfrac{15}{20}$.

   $\frac{(x-2)(x+2)}{(x-2)(x-3)} = \frac{x+2}{x-3}$
4. 4
   That's fully simplified. Note the implicit restriction $x \neq 2$ (or we'd have divided by zero).

   $\boxed{\dfrac{x+2}{x-3}}$

Worked Example: Solving a Quadratic with the Quadratic Formula

1. 1
   Identify the coefficients. The quadratic is in the form $ax^2 + bx + c = 0$.

   $a = 2, \quad b = 5, \quad c = -4$
2. 2
   Sub into the formula. Watch the signs — wrap that $-4$ in brackets so the $-4ac$ becomes $+32$.

   $x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-4)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 32}}{4}$
3. 3
   Simplify under the root. $\sqrt{57}$ doesn't simplify nicely, so keep as a decimal.

   $x = \frac{-5 \pm \sqrt{57}}{4} = \frac{-5 \pm 7.5498...}{4}$
4. 4
   Two solutions — one with $+$, one with $-$. Round each to 3 s.f. at the end.

   $x = \frac{-5 + 7.5498}{4} = 0.6375 \approx 0.637$ $x = \frac{-5 - 7.5498}{4} = -3.1375 \approx -3.14$

Worked Example: Turning Point via Differentiation

1. 1
   Differentiate term by term using $\frac{d}{dx}(ax^n) = anx^{n-1}$. Constants vanish.

   $\frac{dy}{dx} = 3x^2 - 12x + 9$
2. 2
   Stationary points happen where the gradient is zero. Set the derivative to zero and solve.

   $3x^2 - 12x + 9 = 0$ $x^2 - 4x + 3 = 0$ $(x-1)(x-3) = 0$ $x = 1 \text{ or } x = 3$
3. 3
   Find the $y$-coordinate at each $x$ by subbing back into the original equation.

   $x = 1: y = 1 - 6 + 9 + 2 = 6 \Rightarrow (1, 6)$ $x = 3: y = 27 - 54 + 27 + 2 = 2 \Rightarrow (3, 2)$
4. 4
   Classify by checking gradient just before and after each point. At $x = 1$: gradient at $x=0$ is $+9$ (positive), at $x=2$ is $3(4)-12(2)+9 = -3$ (negative). Sign change $+ \to -$ means maximum. At $x = 3$: gradient at $x=2$ is $-3$, at $x=4$ is $3(16)-12(4)+9 = 9$ (positive). Sign change $- \to +$ means minimum.

   $(1, 6) \text{ is a maximum} \quad ; \quad (3, 2) \text{ is a minimum}$

Worked Example: Composite Function $fg(x)$ and Inverse $f^{-1}(x)$

1. 1
   For (a), $fg(2)$ means $f(g(2))$ — apply $g$ first, then $f$. Read it right-to-left.

   $g(2) = 2^2 - 1 = 3$ $f(3) = 2(3) + 3 = 9$ $fg(2) = 9$
2. 2
   For (b), $gf(x) = g(f(x))$. Sub $f(x) = 2x+3$ into $g$.

   $g(f(x)) = (2x+3)^2 - 1$ $= 4x^2 + 12x + 9 - 1 = 4x^2 + 12x + 8$
3. 3
   For (c), to invert $f$: write $y = 2x + 3$, swap $x$ and $y$, solve for $y$.

   $x = 2y + 3 \implies y = \frac{x - 3}{2}$
4. 4
   Write the inverse using the right notation.

   $f^{-1}(x) = \frac{x - 3}{2}$