18 questions arranged by DBE cognitive level — the parameter \(p\) on parabolas, hyperbolas and exponential graphs, and average gradient. Work each one on paper first, then reveal the memo.
18
practice questions
4
cognitive levels
9
real exam citations
100%
independently verified
How to use this bank.
Start at Level 1 and move up — don't jump to Level 4 first.
Every time you see \(x+p\) (or \(x-p\) on a real paper), solve for where the bracket is zero rather than reading the sign off by eye.
For average gradient questions, find the two points first, then apply \(m=\frac{y_2-y_1}{x_2-x_1}\) — the same formula as straight-line gradient.
Reveal the memo only after a genuine attempt.
Accuracy note: every question below was independently solved from scratch before publication — nine carry a real citation, confirmed against the archived paper and, where available, its official memo; the rest are labelled "Equation Station SA Practice Question."
L1 — Knowledge 4 Qs
L2 — Routine Procedures 5 Qs
L3 — Complex Procedures 4 Qs
L4 — Problem Solving 4 Qs
20%
Level 1 | Knowledge
Direct Asymptote Recall
Read the asymptotes straight off an equation already in standard form.
Q1Equation Station SA Practice Question2 marks
Hyperbola
Asymptotes of a Hyperbola
Write down the equations of the asymptotes of \(f(x)=\dfrac{2}{x+3}-1\).
Turning points, y-intercepts and short algebraic simplification.
Q5Equation Station SA Practice Question4 marks
Parabola
Turning Point With \(p\)
Given \(f(x)=3(x+2)^2-5\), determine the turning point and the equation of the axis of symmetry.
Memo
✓ \(x+2=0\Rightarrow x=-2\)✓ Turning point: \((-2,-5)\)✓ Axis of symmetry: \(x=-2\)
Q6Mpumalanga, November 20251 mark
Exponential
Asymptote of an Exponential With \(p\)
Write down the equation of the asymptote of \(g(x)=2\left(\dfrac12\right)^{x+1}-4\).
Memo
✓ The horizontal asymptote depends only on \(q\): \(y=-4\)
Q7Mpumalanga, November 20251 mark
Parabola
Range of a Parabola With \(p\)
Write down the range of \(f(x)=-(x-3)^2+25\).
Memo
✓ \(a=-1<0\), so the parabola opens downward with maximum value \(25\) at the turning point✓ Range: \(y\le25\)
Q8Equation Station SA Practice Question2 marks
Exponential
Y-Intercept of an Exponential With \(p\)
Given \(h(x)=3^{x+1}-2\), determine the y-intercept of \(h\).
Memo
✓ Let \(x=0\): \(h(0)=3^{0+1}-2=3-2=1\)✓ Y-intercept: \((0,1)\)
Q9Eastern Cape, June 20242 marks
Hyperbola
Simplifying to Standard Form
Given \(f(x)=\dfrac{1}{x-3}-\dfrac{2x+6}{x+3}\), show that \(f(x)\) can be written as \(f(x)=\dfrac{1}{x-3}-2\).
Memo
✓ Factorise the numerator: \(2x+6=2(x+3)\)✓ So \(\dfrac{2x+6}{x+3}=\dfrac{2(x+3)}{x+3}=2\) (for \(x\neq-3\))✓ Therefore \(f(x)=\dfrac{1}{x-3}-2\), as required
30%
Level 3 | Complex Procedures
Multi-Step Methods
Average gradient, axes of symmetry, and transformed hyperbolas.
Q10Gauteng (Johannesburg), June 20235 marks
Average gradient
Average Gradient of a Parabola
For \(g(x)=-\dfrac12x^2+6x-10\), \(B\) is an x-intercept of \(g\) and \(F\) is the turning point of \(g\). Determine the average gradient of \(g\) between \(B\) and \(F\).
Memo
✓ x-intercepts: \(-\dfrac12x^2+6x-10=0\Rightarrow x^2-12x+20=0\Rightarrow(x-10)(x-2)=0\), so \(x=10\) or \(x=2\); take \(B(10,0)\)✓ Turning point: \(x=-\dfrac{b}{2a}=-\dfrac{6}{2(-\frac12)}=6\); \(g(6)=-18+36-10=8\), so \(F(6,8)\)✓ Average gradient: \(m=\dfrac{8-0}{6-10}=\dfrac{8}{-4}=-2\)
Q11Limpopo, June 20252 marks
Average gradient
Average Gradient From Two Given Points
A parabola \(f\) passes through the points \(D(-1,4)\) and \(E(0,6)\). Determine the average gradient of \(f\) between \(D\) and \(E\).
Memo
✓ \(m=\dfrac{6-4}{0-(-1)}=\dfrac{2}{1}=2\)
Q12Limpopo, June 20255 marks
Hyperbola
Rewriting an Improper Rational Function
The graph of \(f(x)=\dfrac{4}{x-2}+4\) is transformed to \(h(x)=\dfrac{x+2}{x-2}\). Determine the horizontal asymptote of \(h\), and describe, in words, the transformation from \(f\) to \(h\).
Memo
✓ Write \(x+2=(x-2)+4\), so \(h(x)=\dfrac{(x-2)+4}{x-2}=1+\dfrac{4}{x-2}=\dfrac{4}{x-2}+1\)✓ Horizontal asymptote: \(y=1\)✓ Comparing \(q\)-values (4 for \(f\), 1 for \(h\)): \(h\) is \(f\) shifted 3 units down
Q13Equation Station SA Practice Question3 marks
Hyperbola
Axis of Symmetry With Positive Gradient
Given \(k(x)=\dfrac{-2}{x+1}+3\), determine the equation of the axis of symmetry of \(k\) that has a positive gradient.
Memo
✓ The centre of the hyperbola is at \((-p,q)=(-1,3)\)✓ The axis of symmetry with gradient \(+1\) passes through the centre: \(y-3=1\cdot(x-(-1))\)✓ \(\boxed{y=x+4}\)
15%
Level 4 | Problem Solving
Full Riders, Multiple Skills Combined
Finding unknown points first, then chaining two or three techniques together.
Q14Mpumalanga, November 20256 marks
Average gradient
Average Gradient Between Two Curves
Given \(f(x)=-(x-3)^2+25\) and \(g(x)=2\left(\dfrac12\right)^{x+1}-4\), where \(A\) is a common x-intercept of \(f\) and \(g\), and \(B\) is the y-intercept of \(g\), determine the average gradient between \(A\) and \(B\).
Memo
✓ Check \(x=-2\): \(f(-2)=-(-5)^2+25=-25+25=0\) — and \(g(-2)=2\left(\dfrac12\right)^{-1}-4=2(2)-4=0\), so \(A(-2,0)\)✓ y-intercept of \(g\): \(g(0)=2\left(\dfrac12\right)^{1}-4=1-4=-3\), so \(B(0,-3)\)✓ Average gradient: \(m=\dfrac{-3-0}{0-(-2)}=-\dfrac32\)
Q15Equation Station SA Practice Question4 marks
Parabola
Full Transformation Description
Describe, in words, the transformation from \(y=x^2\) to \(y=-3(x+2)^2-1\), and hence write down the coordinates of the turning point.
Memo
✓ \(a=-3\): reflect in the x-axis and stretch vertically by a factor of 3✓ \(p=2\): since \(x+2=0\Rightarrow x=-2\), shift 2 units left✓ \(q=-1\): shift 1 unit down✓ Turning point: \(\boxed{(-2,-1)}\)
Q16Eastern Cape, June 20243 marks
Hyperbola
Transformation Between Two Hyperbolas
Given \(f(x)=\dfrac{1}{x-3}-2\), describe, in words, the transformation that maps \(f\) onto \(h(x)=\dfrac{1}{x}\).
Memo
✓ \(f\) has centre \((3,-2)\); \(h\) has centre \((0,0)\)✓ To move from \((3,-2)\) to \((0,0)\): shift 3 units left and 2 units up
Q17Equation Station SA Practice Question6 marks
Hyperbola
Determine \(a\), \(p\) and \(q\) From Given Features
A hyperbola \(f(x)=\dfrac{a}{x+p}+q\) has asymptotes \(x=-1\) and \(y=2\), and passes through the point \((1,4)\). Determine the values of \(p\), \(q\) and \(a\).
Determine the Equation From a Sketch (No Equation Given)
The graph of \(f(x)=\dfrac{a}{x+p}+q\) is sketched below with its asymptotes and one point labelled (no equation given). Determine the values of \(a\), \(p\) and \(q\).
Memo
✓ Vertical asymptote \(x=1\): solve \(x+p=0\) at that value, so \(1+p=0\Rightarrow p=-1\)✓ Horizontal asymptote \(y=-3\): \(q=-3\)✓ Substitute \((3,-1)\): \(-1=\dfrac{a}{3-1}-3\Rightarrow2=\dfrac{a}{2}\Rightarrow a=4\)✓ Check: \(f(3)=\dfrac{4}{3-1}-3=2-3=-1\) ✓, giving \(f(x)=\dfrac{4}{x-1}-3\)