Number & Algebra

Functions and Relationships

Year 10

✓Identify and distinguish between linear, quadratic, and exponential functions based on their equations and graphs.
✓Graph linear, quadratic, and exponential functions, accurately identifying and labelling key features such as intercepts, gradient, vertex, and asymptotes.
✓Solve simultaneous linear equations using algebraic methods (substitution and elimination) and graphical methods.
✓Solve simultaneous equations involving a linear function and a quadratic function algebraically and interpret the solutions graphically.
✓Apply knowledge of functions and simultaneous equations to solve practical problems in various contexts.

Key concepts

Linear Functions

A linear function is a relationship between two variables where the graph is a straight line. The rate of change (gradient) is constant. The general form is y = mx + c, where 'm' represents the gradient and 'c' represents the y-intercept (the point where the line crosses the y-axis).

y = mx + c

Quadratic Functions

A quadratic function is a relationship where the highest power of the independent variable (x) is 2. Its graph is a parabola, which is a U-shaped curve. The general form is y = ax^2 + bx + c. The sign of 'a' determines if the parabola opens upwards (a > 0) or downwards (a < 0). The 'c' value is the y-intercept.

y = ax^2 + bx + c

Exponential Functions

An exponential function is a relationship where the independent variable (x) appears as an exponent. These functions describe situations of rapid growth or decay. The general form is y = ab^x. 'a' is the initial value (the y-intercept when x=0), and 'b' is the base or growth/decay factor. If b > 1, it represents growth; if 0 < b < 1, it represents decay. Exponential functions have a horizontal asymptote.

y = ab^x

Simultaneous Linear Equations

A system of simultaneous linear equations consists of two or more linear equations with the same variables. The solution to such a system is the point (or points) where the graphs of all equations intersect. This point satisfies all equations simultaneously. Common algebraic methods for solving are substitution and elimination.

Simultaneous Equations (Linear & Quadratic)

This involves a system where one equation is linear and the other is quadratic. The solutions are the points of intersection between the straight line and the parabola. There can be zero, one (tangent), or two points of intersection. These systems are typically solved using the substitution method.

Key facts to remember

1The gradient 'm' in y = mx + c determines the steepness and direction of a linear function.
2The vertex of a parabola y = ax^2 + bx + c is located at x = -b / (2a).
3Exponential functions y = ab^x have a horizontal asymptote, which is a line the graph approaches but never touches.
4Solutions to simultaneous equations represent the point(s) where their graphs intersect.
5To find x-intercepts of any function, set y = 0 and solve for x.
6To find y-intercepts of any function, set x = 0 and solve for y.
7A system of linear and quadratic equations can have 0, 1, or 2 solutions (intersections).

Worked examples

Example 1

Graph the quadratic function y = x^2 - 2x - 3 and identify its x-intercepts, y-intercept, and vertex.

I1. Find the y-intercept: Set x = 0. y = (0)^2 - 2(0) - 3 = -3. So, the y-intercept is (0, -3).

II2. Find the x-intercepts: Set y = 0. x^2 - 2x - 3 = 0. Factorise: (x - 3)(x + 1) = 0. So, x = 3 or x = -1. The x-intercepts are (3, 0) and (-1, 0).

III3. Find the vertex: The x-coordinate of the vertex is given by x = -b / (2a). For y = x^2 - 2x - 3, a = 1, b = -2. So, x = -(-2) / (2 * 1) = 2 / 2 = 1.

IV4. Find the y-coordinate of the vertex: Substitute x = 1 into the equation. y = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4. So, the vertex is (1, -4).

V5. Plot the intercepts and vertex, then sketch a smooth parabola through these points, opening upwards as a > 0.

Answer

Y-intercept: (0, -3)\nX-intercepts: (3, 0) and (-1, 0)\nVertex: (1, -4)

Remember that the axis of symmetry passes through the vertex.

Example 2

Solve the following system of simultaneous linear equations using the elimination method:\nEquation 1: 3x + 2y = 13\nEquation 2: 2x - 2y = 2

I1. Notice that the 'y' coefficients are +2 and -2. Adding the two equations will eliminate 'y'.\n (3x + 2y) + (2x - 2y) = 13 + 2\n 5x = 15

II2. Solve for 'x':\n x = 15 / 5\n x = 3

III3. Substitute the value of x = 3 into either Equation 1 or Equation 2 to find 'y'. Using Equation 1:\n 3(3) + 2y = 13\n 9 + 2y = 13

IV4. Solve for 'y':\n 2y = 13 - 9\n 2y = 4\n y = 4 / 2\n y = 2

V5. Check the solution (x=3, y=2) in the other equation (Equation 2):\n 2(3) - 2(2) = 6 - 4 = 2. This matches the right side of Equation 2, so the solution is correct.

Answer

x = 3, y = 2 (or the point (3, 2))

The elimination method is often efficient when coefficients of one variable are opposites or can be easily made opposites.

Example 3

Find the points of intersection for the linear function y = x + 2 and the quadratic function y = x^2 - 2x + 4.

I1. Set the two expressions for 'y' equal to each other, as the points of intersection have the same x and y values for both functions:\n x + 2 = x^2 - 2x + 4

II2. Rearrange the equation into the standard quadratic form (ax^2 + bx + c = 0):\n 0 = x^2 - 2x - x + 4 - 2\n 0 = x^2 - 3x + 2

III3. Solve the quadratic equation for 'x'. Factorise the quadratic:\n (x - 1)(x - 2) = 0\n So, x = 1 or x = 2.

IV4. Substitute each 'x' value back into the linear equation (y = x + 2) to find the corresponding 'y' values:\n For x = 1: y = 1 + 2 = 3. Point 1: (1, 3).\n For x = 2: y = 2 + 2 = 4. Point 2: (2, 4).

V5. The points of intersection are (1, 3) and (2, 4).

Answer

The points of intersection are (1, 3) and (2, 4).

Always substitute back into the simpler (linear) equation to minimise calculation errors.

Common mistakes

✗Incorrectly calculating the gradient or y-intercept when graphing linear functions.
✗Making algebraic errors (especially sign errors) when rearranging equations or solving simultaneous systems.
✗Confusing the characteristics of different function types, e.g., thinking an exponential graph is linear.
✗Failing to find both x and y coordinates for points of intersection in simultaneous equations.
✗Not checking solutions by substituting them back into the original equations.

Exam tips

★Always show all your working steps clearly, especially for algebraic solutions, as partial marks are often awarded.
★Sketch graphs whenever possible, even if not explicitly asked, to help visualise the problem and check the reasonableness of your answers.
★For quadratic equations, remember the quadratic formula (x = [-b ± sqrt(b^2 - 4ac)] / 2a) if factorising is difficult or impossible.
★Carefully read the question to ensure you are finding all required features (e.g., both x and y for intercepts, or all points of intersection).

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