Parabolas and Hyperbolas: The Mistakes That Actually Cost Marks

Parabolas and hyperbolas both show up in Grade 11 and Grade 12 Functions and Graphs, often on the same exam, sometimes in the same question comparing two graphs on one set of axes. They share some vocabulary, intercepts, turning point, asymptote, equation, but the two graphs behave differently enough that a method that works for one will quietly fail on the other. The mistakes below are the ones that repeat across NSC papers year after year.

Mistake 1: Looking for a Turning Point on a Hyperbola

A parabola has exactly one turning point, a single highest or lowest value the function actually reaches. A hyperbola has no turning point at all. It has two branches that get closer and closer to its asymptotes without ever touching them, and there is no single 'top' or 'bottom' value to find.

This mistake shows up when a learner sees a curved graph, assumes 'curved means find the turning point', and tries to set the derivative or the standard turning point formula to work on a hyperbola. It doesn't apply, because there is nothing there to find.

• Worked example: A hyperbola has equation y = 2/(x − 1) + 3. Determine the coordinates of its turning point.
• The correct response is that this graph has no turning point, that concept only applies to parabolas (and exponential graphs, which have none either). What the question is very likely actually testing, given the phrasing, is whether you can identify that a hyperbola has none, or it means to ask for the asymptotes instead, which are x = 1 and y = 3.
• The check to build: before reaching for a turning point calculation, glance at the equation. If it's in the form y = a(x + p)² + q, it's a parabola and a turning point exists. If it's in the form y = a/(x + p) + q, it's a hyperbola and the question is asking about asymptotes, not a turning point.

Mistake 2: Finding Only One Asymptote on a Hyperbola

A hyperbola in the standard CAPS form y = a/(x + p) + q has two asymptotes, a vertical one and a horizontal one, and both are required for full marks whenever a question asks for 'the asymptotes' (plural). Learners who are used to exponential graphs, which have only one horizontal asymptote, sometimes carry that single-asymptote habit into hyperbola questions and stop after finding just one.

• Worked example: Determine the equations of the asymptotes of h(x) = 3/(x + 2) − 1.
• The vertical asymptote comes from the value of x that makes the denominator zero: x + 2 = 0, so x = −2.
• The horizontal asymptote is the value the graph approaches as x gets very large in either direction, which is simply the constant added outside the fraction: y = −1.
• Full answer: x = −2 and y = −1. A common partial-credit mistake is giving only x = −2, having correctly spotted the vertical asymptote but forgetting the horizontal one exists too.

Mistake 3: Reading Off the Wrong Sign for p in the Hyperbola Formula

In the form y = a/(x + p) + q, the vertical asymptote is x = −p, not x = p. This sign flip is a frequent source of small, avoidable errors. A learner correctly memorises 'the vertical asymptote comes from the denominator' but then reads the value of p straight off the equation without applying the sign change.

• Worked example: A hyperbola has a vertical asymptote at x = 4. Write it in the form y = a/(x + p) + q.
• Since the vertical asymptote is x = −p, and here x = 4, that means −p = 4, so p = −4.
• The equation form becomes y = a/(x − 4) + q, not y = a/(x + 4) + q. Writing '+ 4' here, without flipping the sign, is the single most common error in this section.

Mistake 4: Assuming a Parabola's Turning Point Is Always at the Origin or Always a Minimum

In the form y = a(x + p)² + q, the turning point is at (−p ; q), same sign flip on p as the hyperbola's asymptote. Whether that turning point is a minimum or a maximum depends entirely on the sign of a: positive a means the parabola opens upward and the turning point is a minimum, negative a means it opens downward and the turning point is a maximum. Learners who skip checking the sign of a sometimes state the wrong type, or assume the range is 'y ≥ q' when it should be 'y ≤ q'.

• Worked example: f(x) = −2(x − 3)² + 5. Determine the turning point and the range of f.
• Turning point: using (−p ; q) with the equation written as (x − 3), that's (x + (−3)), so p = −3 and the turning point is at (3 ; 5).
• Since a = −2 is negative, the parabola opens downward, so (3 ; 5) is a maximum, and the range is y ≤ 5, not y ≥ 5.
• Getting the coordinate right but the direction of the inequality wrong is a very common way to lose the final mark on this exact question type.

Mistake 5: Solving x-Intercepts on a Hyperbola the Same Way as a Parabola

To find x-intercepts on either graph, the underlying instruction is the same: set y = 0 and solve for x. What differs is the algebra required to get there. A parabola x-intercept question usually means factorising a quadratic or using the quadratic formula. A hyperbola x-intercept question means setting the fraction equal to zero, which only happens when the numerator is zero (a fraction is zero exactly when its top is zero, never because of the bottom).

• Worked example: Determine the x-intercept of g(x) = 5/(x + 2) − 1.
• Set g(x) = 0: 5/(x + 2) − 1 = 0, so 5/(x + 2) = 1.
• Multiply both sides by (x + 2): 5 = x + 2, so x = 3. The x-intercept is (3 ; 0).
• A mistake here is trying to 'cross-multiply' before isolating the fraction, or attempting to factorise as if it were a quadratic. There is no squared term, so there is nothing to factorise, only a fraction to clear.

Mistake 6: Mixing Up Which Graph a Given Set of Clues Describes

Grade 11 and Grade 12 papers frequently give a description in words, an intercept, an asymptote, a turning point, and ask you to write down the equation, without first telling you whether it's a parabola or a hyperbola. The clue is usually hidden in which features are mentioned: a turning point only makes sense for a parabola, and asymptotes only make sense for a hyperbola (or exponential). Reading the clues correctly before choosing a formula avoids setting up the wrong equation entirely.

• Worked example: A graph has asymptotes at x = 1 and y = −2, and passes through the point (3 ; 0). Determine the equation.
• Asymptotes mentioned, no turning point mentioned, this is a hyperbola. Use the form y = a/(x + p) + q.
• Vertical asymptote x = 1 means −p = 1, so p = −1. Horizontal asymptote y = −2 means q = −2. So far: y = a/(x − 1) − 2.
• Substitute the given point (3 ; 0) to solve for a: 0 = a/(3 − 1) − 2, so a/2 = 2, so a = 4. Final equation: y = 4/(x − 1) − 2.
• A learner who misreads this as a parabola question, because 'determine the equation' sounds generic, will set up y = a(x + p)² + q instead and get stuck, since there's no turning point given to find p and q from.

A Quick Way to Sort Parabola Questions From Hyperbola Questions

Before calculating anything, check which words and features appear in the question. This takes a few seconds and prevents the most common mistake on this topic, applying the wrong formula family from the start.

1. Turning point, minimum, maximum mentioned: parabola. Use y = a(x + p)² + q.
2. Asymptote (vertical and horizontal) mentioned: hyperbola. Use y = a/(x + p) + q.
3. x² appears in the equation: parabola.
4. x appears only in a denominator: hyperbola.
5. Once you know which one you're working with, remember the shared sign flip: the value inside the bracket or denominator is the negative of the coordinate or asymptote it produces.