Functions and graphs is the one topic in Grade 12 Maths that connects to almost everything else in Paper 1. It carries roughly 35 marks, and if you cannot sketch, interpret, or read a graph properly, you will lose marks in calculus, in equations, and even in some inequality questions. The good news is that once you understand how the four main graph types work, this topic becomes one of the most predictable in the NSC exam.
In This Post You Will Learn
✓ The four graph types you must know for Grade 12 and how they differ from Grade 11 versions
✓ How to find equations from given graphs using key points
✓ How to determine the domain, range, asymptotes, and axes of symmetry for each function
✓ Step-by-step methods for sketching each graph type from its equation
✓ How to answer "read from the graph" questions that carry easy marks
✓ Where this topic appears in the NSC exam and how many marks it carries
The Four Function Types in Grade 12 Maths
In the CAPS curriculum for Grade 12, you work with four types of functions. You already met these in Grade 10 and 11, but now the questions are harder and the graphs are combined with other topics.
Here is a summary of the four types:
| Function Type | General Form | Key Features |
|---|---|---|
| Parabola (quadratic) | y = a(x - p)² + q | Turning point, axis of symmetry, two x-intercepts possible |
| Hyperbola | y = a/(x - p) + q | Two asymptotes, two curves, no turning point |
| Exponential | y = a.b^(x - p) + q | One asymptote, rapid growth or decay, no x-intercept sometimes |
| Logarithmic | y = log_b(x) | Inverse of exponential, vertical asymptote at x = 0 |
The logarithmic function is the new one in Grade 12. It is the inverse of the exponential function, and questions often ask you to sketch a function and then its inverse on the same set of axes.
How to Sketch a Parabola Step by Step
The standard form you will work with most often is y = a(x - p)² + q.
Here is how to sketch it every single time:
Step 1: Find the Turning Point
It is always at (p, q). If the equation is y = 2(x - 3)² + 1, the turning point is (3, 1).
Step 2: Determine the Shape
If a > 0, the parabola opens upward (happy face). If a < 0, it opens downward (sad face).
Step 3: Find the y-intercept
Set x = 0 and solve for y.
For y = 2(x - 3)² + 1:
y = 2(0 - 3)² + 1
y = 2(9) + 1
y = 19
So the y-intercept is (0, 19).
Step 4: Find the x-intercepts
Set y = 0 and solve for x.
0 = 2(x - 3)² + 1
-1 = 2(x - 3)²
-1/2 = (x - 3)²
You cannot square root a negative number, so this parabola has no x-intercepts. It sits entirely above the x-axis.
Step 5: Draw the Axis of Symmetry
This is the vertical line x = p. In our example, x = 3.
Step 6: Plot and Connect
Plot the turning point, y-intercept, and the mirror image of the y-intercept across the axis of symmetry. Connect with a smooth curve.
Converting From y = ax² + bx + c to Turning Point Form
Sometimes the equation is not in turning point form. If you get y = 2x² - 12x + 19, you need to find the turning point another way.
Use the formula: x = -b/(2a) to find the x-value of the turning point. Then substitute back to find y.
x = -(-12)/(2 x 2) = 12/4 = 3
y = 2(3)² - 12(3) + 19 = 18 - 36 + 19 = 1
Turning point is (3, 1). Same answer as before.
How to Sketch a Hyperbola Step by Step
The general form is y = a/(x - p) + q.
The two asymptotes are your starting point for every hyperbola sketch.
Vertical asymptote: x = p (the graph never touches this line)
Horizontal asymptote: y = q (the graph approaches but never reaches this value)
Example: Sketch y = 3/(x - 2) + 1
Step 1: Asymptotes are x = 2 and y = 1. Draw these as dashed lines first.
Step 2: Since a = 3 (positive), the curves sit in the first and third quadrants relative to the intersection of the asymptotes.
Step 3: Find the y-intercept. Set x = 0:
y = 3/(0 - 2) + 1 = -3/2 + 1 = -1/2
Step 4: Find the x-intercept. Set y = 0:
0 = 3/(x - 2) + 1
-1 = 3/(x - 2)
x - 2 = -3
x = -1
Step 5: Plot the intercepts and draw two smooth curves approaching the asymptotes.
The domain is x ∈ R, x ≠ 2. The range is y ∈ R, y ≠ 1.
The Exponential Function and Its Inverse
The exponential function has the form y = a.b^(x - p) + q where b > 0 and b ≠ 1.
The horizontal asymptote is y = q. If b > 1, the graph increases. If 0 < b < 1, the graph decreases.
Example: y = 2^x
This passes through (0, 1) because 2^0 = 1. The asymptote is y = 0. The graph increases as x gets larger.
How to Find the Inverse (the Logarithmic Function)
To find the inverse of y = 2^x:
Step 1: Swap x and y to get x = 2^y
Step 2: Solve for y using the definition of a logarithm: y = log₂(x)
The inverse of an exponential function is a logarithmic function. When you sketch these on the same axes, they are reflections of each other in the line y = x.
Key facts about y = log₂(x):
The domain is x > 0 (you cannot take the log of zero or a negative number).
The x-intercept is (1, 0) because log₂(1) = 0.
The vertical asymptote is x = 0.
For full live lessons on this topic, see our Grade 12 Maths tuition page.
How to Find the Equation of a Graph From Given Points
This is one of the most common question types in the NSC exam. You are given a graph with some key points labelled and you must determine the equation.
Finding a Parabola Equation From the Turning Point
If the turning point is given, use y = a(x - p)² + q. Substitute the turning point for p and q, then use another point on the graph to solve for a.
Example: A parabola has turning point (1, -4) and passes through (3, 0).
y = a(x - 1)² - 4
Substitute (3, 0):
0 = a(3 - 1)² - 4
0 = 4a - 4
a = 1
The equation is y = (x - 1)² - 4.
Finding a Hyperbola Equation From Asymptotes
If the asymptotes are given, you know p and q. Use an intercept or other given point to find a.
Finding an Exponential Equation
If you are told the base (e.g. base 2), substitute a known point to find a.
Reading Information From Grade 12 Functions and Graphs
The NSC exam loves asking you to read information directly from graphs. These are often the easier marks, but students lose them because they rush.
Common "read from graph" questions:
"For which values of x is f(x) > 0?" This means: where is the graph above the x-axis? Look at the graph, find where it is above the line y = 0, and give the x-values as an interval.
"For which values of x is f(x) > g(x)?" This means: where is f higher than g? Find the intersection points of f and g, then determine which graph is on top between those points.
"Determine the value of x for which f(x) = g(x)." This means: where do the graphs cross? Read the x-values of the intersection points.
"Write down the range of f." This requires you to look at the lowest and highest y-values the graph reaches. For a parabola opening upward with turning point (1, -4), the range is y ≥ -4 or y ∈ [-4, ∞).
Understanding Domain and Range for Every Function Type
Every function question in the exam will test whether you understand domain and range.
Domain = all the possible x-values (input values)
Range = all the possible y-values (output values)
Quick reference:
| Function | Domain | Range |
|---|---|---|
| Parabola (a > 0, TP at q) | x ∈ R | y ≥ q |
| Parabola (a < 0, TP at q) | x ∈ R | y ≤ q |
| Hyperbola (asymptotes x = p, y = q) | x ∈ R, x ≠ p | y ∈ R, y ≠ q |
| Exponential (asymptote y = q, a > 0) | x ∈ R | y > q |
| Logarithmic (base > 1) | x > 0 | y ∈ R |
Common Mistakes Students Make
- Mixing up the turning point signs
When the equation is y = 2(x - 3)² + 1, the turning point is (3, 1), not (-3, 1). Students see the minus sign in front of the 3 and write -3 as the x-coordinate. Remember: the form is (x - p), so if you see (x - 3), then p = 3. If you see (x + 3), rewrite it as (x - (-3)), so p = -3.
- Forgetting to state restrictions on domain and range
For a hyperbola with asymptotes at x = 2 and y = 1, the domain is not just "all real numbers." It is x ∈ R, x ≠ 2. If you forget the restriction, you lose the mark.
- Drawing asymptotes as solid lines
Asymptotes must always be drawn as dashed lines. A solid line means the graph touches that line. Dashed means it approaches but never reaches. Examiners are strict about this.
- Not finding enough points before sketching
Some students plot the turning point and one intercept, then guess the rest. Always find the y-intercept, both x-intercepts (if they exist), and use the axis of symmetry to mirror points. The more points you plot, the more accurate your sketch.
- Confusing the inverse with the negative
The inverse of y = 2^x is y = log₂(x), not y = -2^x. The negative reflects the graph in the x-axis. The inverse reflects it in the line y = x. These are completely different transformations.
How This Topic Appears in the NSC Exam
Functions and graphs appears in Paper 1 of the NSC Maths exam.
It typically carries between 30 and 40 marks, making it one of the two biggest topics in Paper 1 alongside calculus.
In recent years, this topic has consistently appeared as one of the first major questions, usually around Question 3 to Question 5. The DBE tends to structure it in two or three sub-questions.
One sub-question gives you equations and asks you to sketch the graphs, label key points, and state domain and range. Another sub-question gives you graphs and asks you to determine the equations. A third part usually tests reading from graphs: inequalities, intersections, increasing/decreasing intervals.
In the 2023 NSC exam, functions appeared as a combined question involving a parabola and a straight line, where students had to find intersection points, determine the equation of the parabola, and answer inequality questions based on the graphs.
The inverse function (logarithmic) has appeared every year since it was introduced to the CAPS syllabus. Expect at least one question asking you to sketch a function and its inverse on the same axes, or to write down the equation of the inverse.
The pattern is clear: if you can sketch all four graph types, find equations from graphs, and answer "read from graph" inequality questions, you are looking at 30+ marks in Paper 1.
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