Why This Matters for Grade 11#
In Grade 11, every function gains a horizontal shift parameter $p$. But that only makes sense if you already understand:
- What domain and range mean and how to write them
- How the parameter $a$ controls shape and reflection
- How the parameter $q$ controls the vertical shift
- How to find intercepts and asymptotes
If these basics aren’t automatic, the Grade 11 parabola, hyperbola, and exponential will feel overwhelming.
1. What is a Function?#
A function is a rule that assigns exactly one output to each input.
- Input = $x$ (independent variable)
- Output = $y$ or $f(x)$ (dependent variable)
Function notation: $f(x) = 2x + 3$ means “the function $f$ takes $x$ and returns $2x + 3$”.
$f(4) = 2(4) + 3 = 11$ → “the value of $f$ at $x = 4$ is $11$”
Key: $f(x)$ does NOT mean “$f$ times $x$”. It means “the function $f$ applied to $x$”.
2. Domain and Range#
| Term | Meaning | How to find it |
|---|---|---|
| Domain | All possible $x$-values (inputs) | Look for restrictions: division by zero, square roots of negatives |
| Range | All possible $y$-values (outputs) | Look at the graph — what $y$-values does it actually reach? |
Examples from Grade 10#
| Function | Domain | Range |
|---|---|---|
| $y = 2x + 3$ (linear) | $x \in \mathbb{R}$ | $y \in \mathbb{R}$ |
| $y = x^2 + 1$ (parabola, $a > 0$) | $x \in \mathbb{R}$ | $y \geq 1$ |
| $y = -x^2 + 4$ (parabola, $a < 0$) | $x \in \mathbb{R}$ | $y \leq 4$ |
| $y = \frac{3}{x}$ (hyperbola) | $x \in \mathbb{R}, x \neq 0$ | $y \in \mathbb{R}, y \neq 0$ |
| $y = 2^x + 1$ (exponential) | $x \in \mathbb{R}$ | $y > 1$ |
Grade 11 upgrade: When the horizontal shift $p$ is added, the domain restriction and asymptote shift too.
3. The Effect of $a$ and $q$#
In Grade 10, every function has the form with parameters $a$ and $q$:
| Parameter | Effect | Example |
|---|---|---|
| $a > 0$ | Graph opens upward / sits in standard position | $y = 2x^2$ opens up |
| $a < 0$ | Graph reflects (flips) | $y = -2x^2$ opens down |
| $\|a\| > 1$ | Graph is stretched (narrower) | $y = 3x^2$ is narrower than $y = x^2$ |
| $0 < \|a\| < 1$ | Graph is compressed (wider) | $y = \frac{1}{2}x^2$ is wider than $y = x^2$ |
| $q > 0$ | Graph shifts up by $q$ units | $y = x^2 + 3$ shifts up 3 |
| $q < 0$ | Graph shifts down by $\|q\|$ units | $y = x^2 - 2$ shifts down 2 |
4. Finding Intercepts#
$y$-intercept#
Set $x = 0$ and solve for $y$.
For $y = 2x^2 - 8$: $y = 2(0)^2 - 8 = -8$. So the $y$-intercept is $(0; -8)$.
$x$-intercept(s)#
Set $y = 0$ and solve for $x$.
For $y = 2x^2 - 8$: $0 = 2x^2 - 8$, so $x^2 = 4$, $x = \pm 2$. The $x$-intercepts are $(-2; 0)$ and $(2; 0)$.
Key connection: Finding $x$-intercepts = solving $f(x) = 0$. This is why factorisation and equation solving are so important for functions.
5. Asymptotes#
An asymptote is a line that the graph approaches but never touches.
| Function type | Asymptote |
|---|---|
| $y = \frac{a}{x} + q$ (hyperbola) | Horizontal: $y = q$, Vertical: $x = 0$ |
| $y = ab^x + q$ (exponential) | Horizontal: $y = q$ |
In Grade 11, the vertical asymptote shifts to $x = p$ and the horizontal asymptote shifts to $y = q$ when the function becomes $y = \frac{a}{x-p} + q$.
6. Reading a Graph#
Given a graph, you should be able to identify:
- Type: Linear, parabola, hyperbola, or exponential
- Intercepts: Where the graph crosses the axes
- Turning point (parabola): The minimum or maximum point — at $(0; q)$ in Grade 10
- Asymptotes (hyperbola/exponential): The lines the graph approaches
- Domain & range: Read from the graph’s extent and restrictions
- Increasing/decreasing: Where the graph goes up vs down
Grade 11 exam tip: “Determine the equation from the graph” questions are very common. You read the key features, then work backwards to find $a$, $p$, and $q$.
🚨 Common Mistakes#
- Confusing $f(x)$ with $f \times x$: $f(3)$ means substitute $x = 3$ into the function, not multiply.
- Range of parabola: $y = -x^2 + 5$ has range $y \leq 5$, not $y \geq 5$. Check if $a$ is positive or negative.
- Asymptote is NOT touched: The graph gets infinitely close but never reaches the asymptote. Don’t draw the curve touching it.
- $x$-intercept when there isn’t one: $y = x^2 + 4$ has no $x$-intercepts because $x^2 + 4 = 0$ has no real solutions. The parabola sits entirely above the $x$-axis.
- Domain of hyperbola: $y = \frac{3}{x}$ has domain $x \neq 0$, not “all real numbers”. The graph has two separate branches.